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Partial Derivatives

2.5 Tangent Planes and Normal Lines

19 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Is it reasonable to say that the surfaces x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1 and x2+y2+(z+1)2=1x^2+y^2+(z+1)^2=1 are tangent to each other at (0,0,0)(0,0,0)?

Hint

What are the tangent planes to the two surfaces at (0,0,0)(0,0,0)?

Answer

Yes. The plane z=0z=0 is the tangent plane to both surfaces at (0,0,0)(0,0,0).

Full solution

Write F(x,y,z)=x2+y2+(z−1)2−1F(x,y,z) = x^2+y^2+(z-1)^2-1 and G(x,y,z)=x2+y2+(z+1)2−1G(x,y,z) = x^2+y^2+(z+1)^2-1. Let S1S_1 denote the surface F(x,y,z)=0F(x,y,z)=0 and S2S_2 denote the surface G(x,y,z)=0G(x,y,z)=0. First note that F(0,0,0)=G(0,0,0)=0F(0,0,0)=G(0,0,0)=0 so that the point (0,0,0)(0,0,0) lies on both S1S_1 and S2S_2. The gradients of FF and GG are

∇F(x,y,z)=<∂F∂x(x,y,z) , ∂F∂y(x,y,z) , ∂F∂z(x,y,z)>=<2x , 2y , 2(z−1)>∇G(x,y,z)=<∂G∂x(x,y,z) , ∂G∂y(x,y,z) , ∂G∂z(x,y,z)>=<2x , 2y , 2(z+1)>\begin{align*} \vnabla F(x,y,z) &=\llt\pdiff{F}{x}(x,y,z)\,,\, \pdiff{F}{y}(x,y,z)\,,\, \pdiff{F}{z}(x,y,z)\rgt =\llt 2x\,,\,2y\,,\,2(z-1)\rgt \\ \vnabla G(x,y,z) &=\llt\pdiff{G}{x}(x,y,z)\,,\, \pdiff{G}{y}(x,y,z)\,,\, \pdiff{G}{z}(x,y,z)\rgt =\llt 2x\,,\,2y\,,\,2(z+1)\rgt \end{align*}

In particular,

∇F(0,0,0)=<0,0,−2>∇G(0,0,0)=<0,0,2>\begin{equation*} \vnabla F(0,0,0)=\llt 0,0,-2\rgt\qquad \vnabla G(0,0,0)=\llt 0,0,2\rgt \end{equation*}

so that the vector k^=−12∇F(0,0,0)=12∇G(0,0,0)\hk=-\frac{1}{2}\vnabla F(0,0,0) =\frac{1}{2}\vnabla G(0,0,0) is normal to both surfaces at (0,0,0)(0,0,0). So the tangent plane to both S1S_1 and S2S_2 at (0,0,0)(0,0,0) is

k^⋅<x−0,y−0,z−0>=0orz=0\begin{equation*} \hk\cdot\llt x-0,y-0,z-0\rgt=0\qquad\text{or}\qquad z=0 \end{equation*}

Denote by PP the plane z=0z=0. Thus S1S_1 is tangent to PP at (0,0,0)(0,0,0) and PP is tangent to S2S_2 at (0,0,0)(0,0,0). So it is reasonable to say that S1S_1 and S2S_2 are tangent at (0,0,0)(0,0,0).

Q2Stage 1

Let the point r0=(x0,y0,z0)\vr_0= (x_0,y_0,z_0) lie on the surface G(x,y,z)=0G(x,y,z)=0. Assume that ∇G(x0,y0,z0)≠0\vnabla G(x_0,y_0,z_0)\ne\vZero. Suppose that the parametrized curve r(t)=(x(t),y(t),z(t))\vr(t)=\big(x(t),y(t),z(t)\big) is contained in the surface and that r(t0)=r0\vr(t_0)=\vr_0. Show that the tangent line to the curve at r0\vr_0 lies in the tangent plane to G=0G=0 at r0\vr_0.

Hint

Apply the chain rule to G(r(t))=0G\big(\vr(t)\big)=0.

Answer

See the solution.

Full solution

Denote by SS the surface G(x,y,z)=0G(x,y,z)=0 and by CC the parametrized curve r(t)=(x(t),y(t),z(t))\vr(t)=\big(x(t),y(t),z(t)\big). To start, we'll find the tangent plane to SS at r0\vr_0 and the tangent line to CC at r0\vr_0.

  • The tangent vector to CC at r0\vr_0 is <x′(t0) , y′(t0) , z′(t0)>\llt x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \rgt, so the parametric equations for the tangent line to CC at r0\vr_0 are

    x−x0=tx′(t0)y−y0=ty′(t0)z−z0=tz′(t0)\begin{equation*} x-x_0 = t x'(t_0)\qquad y-y_0 = t y'(t_0)\qquad z-z_0 = t z'(t_0) \tag{$E_1$}\end{equation*}

  • The gradient <∂G∂x(x0 , y0 , z0) , ∂G∂y(x0 , y0 , z0) , ∂G∂z(x0 , y0 , z0)>\llt\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\rgt is a normal vector to the surface SS at (x0,y0,z0)(x_0,y_0,z_0). So the tangent plane to the surface SS at (x0,y0,z0)(x_0,y_0,z_0) is

    <∂G∂x(x0 , y0 , z0) , ∂G∂y(x0 , y0 , z0) , ∂G∂z(x0 , y0 , z0)>⋅<x−x0 , y−y0 , z−z0>=0\begin{equation*} \llt\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\rgt \cdot \llt x-x_0\,,\, y-y_0\,,\,z-z_0\rgt = 0 \end{equation*}

    or

    ∂G∂x(x0 , y0 , z0) (x−x0)+∂G∂y(x0 , y0 , z0) (y−y0)+∂G∂z(x0 , y0 , z0) (z−z0)=0\begin{equation*} \pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}\end{equation*}

Next, we'll show that the tangent vector
<x′(t0) , y′(t0) , z′(t0)>\llt x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \rgt to CC at r0\vr_0 and the normal vector <∂G∂x(x0 , y0 , z0) , ∂G∂y(x0 , y0 , z0) , ∂G∂z(x0 , y0 , z0)>\llt\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\rgt to SS at r0\vr_0 are perpendicular to each other. To do so, we observe that, for every tt, the point (x(t),y(t),z(t))\big(x(t),y(t),z(t)\big) lies on the surface G(x,y,z)=0G(x,y,z)=0 and so obeys

G(x(t),y(t),z(t))=0\begin{align*} G\big(x(t),y(t),z(t)\big) =0 \end{align*}

Differentiating this equation with respect to tt gives, by the chain rule,

0=ddtG(x(t),y(t),z(t))=∂G∂x(x(t) , y(t) , z(t)) x′(t)+∂G∂y(x(t) , y(t) , z(t)) y′(t)+∂G∂z(x(t) , y(t) , z(t)) z′(t)\begin{align*} 0&= \diff{}{t}G\big(x(t),y(t),z(t)\big) \\ &=\pdiff{G}{x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\pdiff{G}{y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\pdiff{G}{z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{align*}

Then setting t=t0t=t_0 gives

∂G∂x(x0 , y0 , z0) x′(t0)+∂G∂y(x0 , y0 , z0) y′(t0)+∂G∂z(x0 , y0 , z0) z′(t0)=0\begin{equation*} \pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$} \end{equation*}

Finally, we are in a position to show that if (x,y,z)(x,y,z) is any point on the tangent line to CC at r0\vr_0, then (x,y,z)(x,y,z) is also on the tangent plane to SS at r0\vr_0. As (x,y,z)(x,y,z) is on the tangent line to CC at r0\vr_0 then there is a tt such that, by (E1)(E_1),

∂G∂x(x0 , y0 , z0) {x−x0}+∂G∂y(x0 , y0 , z0) {y−y0}+∂G∂z(x0 , y0 , z0) {z−z0}=∂G∂x(x0 , y0 , z0) {t x′(t0)}+∂G∂y(x0 , y0 , z0) {t y′(t0)}+∂G∂z(x0 , y0 , z0) {t z′(t0)}=t[∂G∂x(x0 , y0 , z0) x′(t0)+∂G∂y(x0 , y0 , z0) y′(t0)+∂G∂z(x0 , y0 , z0) z′(t0)]=0\begin{align*} &\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\{x-x_0\}} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\{y-y_0\}} +\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\{z-z_0\}} \\ &=\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\big\{ t\,x'(t_0)\big\}} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\big\{ t\,y'(t_0)\big\}} +\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{\big\{ t\,z'(t_0)\big\}}\\ &=\textcolor{blue}{t}\left[\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{x'(t_0)} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{y'(t_0)} +\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \textcolor{blue}{z'(t_0)} \right] =0 \end{align*}

by (E3)(E_3). That is, (x,y,z)(x,y,z) obeys the equation, (E2)(E_2), of the tangent plane to SS at r0\vr_0 and so is on that tangent plane. So the tangent line to CC at r0\vr_0 is contained in the tangent plane to SS at r0\vr_0.

Q3Stage 1

Let F(x0,y0,z0)=G(x0,y0,z0)=0F(x_0,y_0,z_0)=G(x_0,y_0,z_0)=0 and let the vectors ∇F(x0,y0,z0)\vnabla F(x_0,y_0,z_0) and ∇G(x0,y0,z0)\vnabla G(x_0,y_0,z_0) be nonzero and not be parallel to each other. Find the equation of the normal plane to the curve of intersection of the surfaces F(x,y,z)=0F(x,y,z)=0 and G(x,y,z)=0G(x,y,z)=0 at (x0,y0,z0)(x_0,y_0,z_0). By definition, that normal plane is the plane through (x0,y0,z0)(x_0,y_0,z_0) whose normal vector is the tangent vector to the curve of intersection at (x0,y0,z0)(x_0,y_0,z_0).

Hint

To find a tangent vector to the curve of intersection of the surfaces F(x,y,z)=0F(x,y,z)=0 and G(x,y,z)=0G(x,y,z)=0 at (x0,y0,z0)(x_0,y_0,z_0), use Q[2] twice, once for the surface F(x,y,z)=0F(x,y,z)=0 and once for the surface G(x,y,z)=0G(x,y,z)=0.

Answer

The normal plane is n⋅<x−x0 , y−y0 , z−z0>=0\vn\cdot\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =0, where the normal vector n=∇F(x0,y0,z0)×∇G(x0,y0,z0)\vn = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0).

Full solution

Use S1S_1 to denote the surface F(x,y,z)=0F(x,y,z)=0, S2S_2 to denote the surface G(x,y,z)=0G(x,y,z)=0 and CC to denote the curve of intersection of S1S_1 and S2S_2.

  • Since CC is contained in S1S_1, the tangent line to CC at (x0,y0,z0)(x_0,y_0,z_0) is contained in the tangent plane to S1S_1 at (x0,y0,z0)(x_0,y_0,z_0), by Q[2]. In particular, any tangent vector, t\vt, to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendicular to ∇F(x0,y0,z0)\vnabla F(x_0,y_0,z_0), the normal vector to S1S_1 at (x0,y0,z0)(x_0,y_0,z_0).

  • Since CC is contained in S2S_2, the tangent line to CC at (x0,y0,z0)(x_0,y_0,z_0) is contained in the tangent plane to S2S_2 at (x0,y0,z0)(x_0,y_0,z_0), by Q[2]. In particular, any tangent vector, t\vt, to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendicular to ∇G(x0,y0,z0)\vnabla G(x_0,y_0,z_0), the normal vector to S2S_2 at (x0,y0,z0)(x_0,y_0,z_0).

So any tangent vector to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendiular to both ∇F(x0,y0,z0)\vnabla F(x_0,y_0,z_0) and ∇G(x0,y0,z0)\vnabla G(x_0,y_0,z_0). One such tangent vector is

t=∇F(x0,y0,z0)×∇G(x0,y0,z0)\begin{align*} \vt = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{align*}

(Because the vectors ∇F(x0,y0,z0)\vnabla F(x_0,y_0,z_0) and ∇G(x0,y0,z0)\vnabla G(x_0,y_0,z_0) are nonzero and not parallel, t\vt is nonzero.) So the normal plane in question passes through (x0,y0,z0)(x_0,y_0,z_0) and has normal vector n=t\vn=\vt. Consquently, the normal plane is

n⋅<x−x0 , y−y0 , z−z0>=0where n=t=∇F(x0,y0,z0)×∇G(x0,y0,z0)\begin{equation*} \vn\cdot\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =0 \qquad\text{where } \vn=\vt=\vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{equation*}
Q4Stage 1

Let f(x0,y0)=g(x0,y0)f(x_0,y_0)=g(x_0,y_0) and let <fx(x0,y0),fy(x0,y0)>≠<gx(x0,y0),gy(x0,y0)>\llt f_x(x_0,y_0), f_y(x_0,y_0)\rgt\ne \llt g_x(x_0,y_0), g_y(x_0,y_0)\rgt. Find the equation of the tangent line to the curve of intersection of the surfaces z=f(x,y)z=f(x,y) and z=g(x,y)z=g(x,y) at (x0 , y0 , z0=f(x0,y0))(x_0\,,\,y_0\,,\,z_0=f(x_0,y_0)).

Hint

To find a tangent vector to the curve of intersection of the surfaces z=f(x,y)z=f(x,y) and z=g(x,y)z=g(x,y) at (x0,y0,z0)(x_0,y_0,z_0), use Q[2] twice, once for the surface z=f(x,y)z=f(x,y) and once for the surface z=g(x,y)z=g(x,y).

Answer

Tangent line is

x=x0+t[gy(x0,y0)−fy(x0,y0)]y=y0+t[fx(x0,y0)−gx(x0,y0)]z=z0+t[fx(x0,y0)gy(x0,y0)−fy(x0,y0)gx(x0,y0)]\begin{align*} x&=x_0+t\big[g_y(x_0,y_0)-f_y(x_0,y_0)\big] \\ y&=y_0+t\big[f_x(x_0,y_0)-g_x(x_0,y_0)\big] \\ z&=z_0+ t\big[f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\big] \end{align*}
Full solution

Use S1S_1 to denote the surface z=f(x,y)z=f(x,y), S2S_2 to denote the surface z=g(x,y)z=g(x,y) and CC to denote the curve of intersection of S1S_1 and S2S_2.

  • Since CC is contained in S1S_1, the tangent line to CC at (x0,y0,z0)(x_0,y_0,z_0) is contained in the tangent plane to S1S_1 at (x0,y0,z0)(x_0,y_0,z_0), by Q[2]. In particular, any tangent vector, t\vt, to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendicular to −fx(x0,y0) ı^−fy(x0,y0) ȷ^+k^-f_x(x_0,y_0)\,\hi -f_y(x_0,y_0)\,\hj+\hk, the normal vector to S1S_1 at (x0,y0,z0)(x_0,y_0,z_0). (See Theorem 2.5.1 in the CLP-3 text.)

  • Since CC is contained in S2S_2, the tangent line to CC at (x0,y0,z0)(x_0,y_0,z_0) is contained in the tangent plane to S2S_2 at (x0,y0,z0)(x_0,y_0,z_0), by Q[2]. In particular, any tangent vector, t\vt, to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendicular to −gx(x0,y0) ı^−gy(x0,y0) ȷ^+k^-g_x(x_0,y_0)\,\hi -g_y(x_0,y_0)\,\hj+\hk, the normal vector to S2S_2 at (x0,y0,z0)(x_0,y_0,z_0).

So any tangent vector to CC at (x0,y0,z0)(x_0,y_0,z_0) must be perpendicular to both of the vectors −fx(x0,y0) ı^−fy(x0,y0) ȷ^+k^-f_x(x_0,y_0)\,\hi-f_y(x_0,y_0)\,\hj+\hk and −gx(x0,y0) ı^−gy(x0,y0) ȷ^+k^-g_x(x_0,y_0)\,\hi -g_y(x_0,y_0)\,\hj+\hk. One such tangent vector is

t=[−fx(x0,y0) ı^−fy(x0,y0) ȷ^+k^]×[−gx(x0,y0) ı^−gy(x0,y0) ȷ^+k^]=det⁡[ı^ȷ^k^−fx(x0,y0)−fy(x0,y0)1−gx(x0,y0)−gy(x0,y0)1]=<gy(x0,y0)−fy(x0,y0) , fx(x0,y0)−gx(x0,y0) , fx(x0,y0)gy(x0,y0)−fy(x0,y0)gx(x0,y0)>\begin{align*} &\vt = \big[-f_x(x_0,y_0)\,\hi - f_y(x_0,y_0)\,\hj+\hk\big]\times \big[-g_x(x_0,y_0)\,\hi - g_y(x_0,y_0)\,\hj+\hk\big] \\ &=\det\left[\begin{matrix} \hi & \hj & \hk \\ -f_x(x_0,y_0) & -f_y(x_0,y_0) & 1 \\ -g_x(x_0,y_0) & -g_y(x_0,y_0) & 1 \end{matrix}\right] \\ &=\llt g_y(x_0,y_0)-f_y(x_0,y_0)\,,\, f_x(x_0,y_0)-g_x(x_0,y_0)\,,\, f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\rgt \end{align*}

So the tangent line in question passes through (x0,y0,z0)(x_0,y_0,z_0) and has direction vector d=t\vd=\vt. Consquently, the tangent line is

<x−x0 , y−y0 , z−z0>=t d\begin{equation*} \llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt = t\,\vd \end{equation*}

or

x=x0+t[gy(x0,y0)−fy(x0,y0)]y=y0+t[fx(x0,y0)−gx(x0,y0)]z=z0+t[fx(x0,y0)gy(x0,y0)−fy(x0,y0)gx(x0,y0)]\begin{align*} x&=x_0+t\big[g_y(x_0,y_0)-f_y(x_0,y_0)\big] \\ y&=y_0+t\big[f_x(x_0,y_0)-g_x(x_0,y_0)\big] \\ z&=z_0+ t\big[f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\big] \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q5Stage 2Past exam · M200 2009A

Let f(x,y)=x2yx4+2y2\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}. Find the tangent plane to the surface z=f(x,y)z = f(x,y) at the point (−1 , 1 , 13)\left( -1\,,\,1\,,\,\frac{1}{3}\right).

Answer

2x+y+9z=22x+y+9z=2

Full solution

We are going to use Theorem 2.5.1 in the CLP-3 text. To do so, we need the first order derivatives of f(x,y)f(x,y) at (x,y)=(−1,1)(x,y)=(-1,1). So we find them first.

fx(x,y)=2xyx4+2y2−x2y(4x3)(x4+2y2)2fx(−1,1)=−23+432=−29fy(x,y)=x2x4+2y2−x2y(4y)(x4+2y2)2fy(−1,1)=13−432=−19\begin{alignat*}{3} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9} \\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{alignat*}

The tangent plane is

z=f(−1,1)+fx(−1,1) (x+1)+fy(−1,1) (y−1)=13−29 (x+1)−19 (y−1)=29−29x−19y\begin{align*} z&=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \\ &=\frac{2}{9}-\frac{2}{9}x-\frac{1}{9}y \end{align*}

or 2x+y+9z=22x+y+9z=2.

Q6Stage 2Past exam · M200 2015D

Find the tangent plane to

27x2+y2+z2+3=9\begin{equation*} \frac{27}{\sqrt{x^2+y^2+z^2+3}}=9 \end{equation*}

at the point (2,1,1)(2, 1, 1).

Answer

2x+y+z=62x+y+z = 6

Full solution

The equation of the given surface is of the form G(x,y,z)=9G(x,y,z)=9 with G(x,y,z)=27x2+y2+z2+3G(x,y,z) =\frac{27}{\sqrt{x^2+y^2+z^2+3}}. So, by Theorem 2.5.5 in the CLP-3 text, a normal vector to the surface at (2,1,1)(2,1,1) is

∇G(2,1,1)=−12 27(x2+y2+z2+3)3/2(2x , 2y , 2z)∣(x,y,z)=(2,1,1)=−<2 , 1 , 1>\begin{align*} \vnabla G(2,1,1) &=-\frac{1}{2}\ \frac{27}{(x^2+y^2+z^2+3)^{3/2}}\big(2x\,,\,2y\,,\,2z\big) \bigg|_{(x,y,z)=(2,1,1)} \\ &=-\llt 2\,,\,1\,,\,1\rgt \end{align*}

and the equation of the tangent plane is

−<2 , 1 , 1>⋅<x−2 , y−1 , z−1>=0or2x+y+z=6\begin{equation*} -\llt 2\,,\,1\,,\,1\rgt\cdot \llt x-2\,,\,y-1\,,\,z-1\rgt=0\qquad\text{or}\qquad 2x+y+z = 6 \end{equation*}
Q7Stage 2

Find the equations of the tangent plane and the normal line to the graph of the specified function at the specified point.

  1. f(x,y)=x2−y2f(x,y)=x^2-y^2 at (−2,1)(-2,1)

  2. f(x,y)=exyf(x,y)=e^{xy} at (2,0)(2,0)

Answer

(a) The tangent plane is 4x+2y+z=−34x+2y+z=-3 and the normal line is <x,y,z>=<−2,1,3>+t<4,2,1>\llt x,y,z\rgt=\llt -2,1,3\rgt+t\llt 4,2,1\rgt.

(b) The tangent plane is 2y−z=−12y-z=-1 and the normal line is <x,y,z>=<2,0,1>+t<0,2,−1>\llt x,y,z\rgt=\llt 2,0,1\rgt+t\llt 0,2,-1\rgt.

Full solution

(a) The specified graph is z=f(x,y)=x2−y2z=f(x,y)=x^2-y^2 or F(x,y,z)=x2−y2−z=0F(x,y,z)=x^2-y^2-z=0. Observe that f(−2,1)=3f(-2,1)=3. The vector

∇F(−2,1,3)=<Fx(x,y,z),Fy(x,y,z),Fz(x,y,z)>∣(x,y,z)=(−2,1,3)=<2x,−2y,−1>∣(x,y,z)=(−2,1,3)=<−4,−2,−1>\begin{align*} \vnabla F(-2,1,3) &= \llt F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\rgt\Big|_{(x,y,z)=(-2,1,3)} \\ &= \llt 2x,-2y,-1\rgt\Big|_{(x,y,z)=(-2,1,3)} \\ &= \llt -4,-2,-1\rgt \end{align*}

is a normal vector to the graph at (−2,1,3)(-2,1,3). So the tangent plane is

−4(x+2)−2(y−1)−(z−3)=0 or 4x+2y+z=−3\begin{equation*} -4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3 \end{equation*}

and the normal line is

<x,y,z>=<−2,1,3>+t<4,2,1>\begin{equation*} \llt x,y,z\rgt=\llt -2,1,3\rgt+t\llt 4,2,1\rgt \end{equation*}

(b) The specified graph is z=f(x,y)=exyz=f(x,y)=e^{xy} or F(x,y,z)=exy−z=0F(x,y,z)=e^{xy}-z=0. Observe that f(2,0)=1f(2,0)=1. The vector

∇F(2,0,1)=<Fx(x,y,z),Fy(x,y,z),Fz(x,y,z)>∣(x,y,z)=(2,0,1)=<yexy,xexy,−1>∣(x,y,z)=(2,0,1)=<0,2,−1>\begin{align*} \vnabla F(2,0,1) &= \llt F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\rgt\Big|_{(x,y,z)=(2,0,1)} \\ &= \llt ye^{xy},xe^{xy},-1\rgt\Big|_{(x,y,z)=(2,0,1)} \\ &= \llt 0,2,-1\rgt \end{align*}

is a normal vector to the graph at (2,0,1)(2,0,1). So the tangent plane is

0(x−2)+2(y−0)−(z−1)=0 or 2y−z=−1\begin{equation*} 0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1 \end{equation*}

and the normal line is

<x,y,z>=<2,0,1>+t<0,2,−1>\begin{equation*} \llt x,y,z\rgt=\llt 2,0,1\rgt+t\llt 0,2,-1\rgt \end{equation*}
Q8Stage 2Past exam · M200 2005D

Consider the surface z=f(x,y)z = f(x,y) defined implicitly by the equation xyz2+y2z3=3+x2xyz^2 + y^2 z^3 = 3 + x^2. Use a 3–dimensional gradient vector to find the equation of the tangent plane to this surface at the point (−1,1,2)(-1, 1, 2). Write your answer in the form z=ax+by+cz = ax + by + c, where aa, bb and cc are constants.

Answer

z=−34x−32y+114z = -\frac{3}{4} x- \frac{3}{2} y + \frac{11}{4}

Full solution

We may use G(x,y,z)=xyz2+y2z3−3−x2=0G(x,y,z) = xyz^2 + y^2 z^3 - 3 - x^2 = 0 as an equation for the surface. Note that (−1,1,2)(-1,1,2) really is on the surface since

G(−1,1,2)=(−1)(1)(2)2+(1)2(2)3−3−(−1)2=−4+8−3−1=0\begin{align*} G(-1,1,2) = (-1)(1)(2)^2 + (1)^2 (2)^3 - 3 - (-1)^2 = -4 + 8 - 3 - 1 =0 \end{align*}

By Theorem 2.5.5 in the CLP-3 text, since

Gx(x,y,z)=yz2−2xGx(−1,1,2)=6Gy(x,y,z)=xz2+2yz3Gy(−1,1,2)=12Gz(x,y,z)=2xyz+3y2z2Gz(−1,1,2)=8\begin{alignat*}{5} G_x(x,y,z)&=yz^2 -2x \qquad & G_x(-1,1,2)&=6 \\ G_y(x,y,z)&=xz^2 +2yz^3 \qquad & G_y(-1,1,2)&=12 \\ G_z(x,y,z)&=2xyz+3y^2z^2 \qquad & G_z(-1,1,2)&=8 \end{alignat*}

one normal vector to the surface at (−1,1,2)(-1,1,2) is ∇G(−1,1,2)=<6 , 12 , 8>\vnabla G(-1,1,2) = \llt 6\,,\,12\,,\,8\rgt and an equation of the tangent plane to the surface at (−1,1,2)(-1,1,2) is

<6 , 12 , 8>⋅<x+1 , y−1 , z−2>=0or6x+12y+8z=22\begin{align*} \llt 6\,,\,12\,,\,8\rgt \cdot \llt x+1\,,\,y-1\,,\,z-2\rgt = 0\qquad\text{or}\qquad 6x+12 y+ 8z = 22 \end{align*}

or

z=−34x−32y+114\begin{equation*} z = -\frac{3}{4} x- \frac{3}{2} y +\frac{11}{4} \end{equation*}
Q9Stage 2Past exam · M200 2008D

A surface is given by

z=x2−2xy+y2.\begin{equation*} z = x^2 - 2xy + y^2 . \end{equation*}

  1. Find the equation of the tangent plane to the surface at x=ax = a, y=2ay = 2a.

  2. For what value of aa is the tangent plane parallel to the plane x−y+z=1x - y + z = 1?

Answer

(a) 2ax−2ay+z=−a22ax -2ay +z = -a^2 (b) a=12a=\frac{1}{2}.

Full solution

(a) The surface is G(x,y,z)=z−x2+2xy−y2=0G(x,y,z)=z-x^2+2xy-y^2=0. When x=ax=a and y=2ay=2a and (x,y,z)(x,y,z) is on the surface, we have z=a2−2(a)(2a)+(2a)2=a2z= a^2-2(a)(2a) +(2a)^2=a^2. So, by Theorem 2.5.5 in the CLP-3 text, a normal vector to this surface at (a,2a,a2)(a,2a,a^2) is

∇G(a,2a,a2)=<−2x+2y , 2x−2y , 1>∣(x,y,z)=(a,2a,a2)=<2a , −2a , 1>\begin{align*} \vnabla G(a,2a,a^2) = \llt -2x+2y\,,\,2x-2y\,,\,1\rgt\Big|_{(x,y,z)=(a,2a,a^2)} = \llt 2a\,,\,-2a\,,\,1\rgt \end{align*}

and the equation of the tangent plane is

<2a , −2a , 1>⋅<x−a , y−2a , z−a2>=0or2ax−2ay+z=−a2\begin{align*} \llt 2a\,,\,-2a\,,\,1\rgt\cdot\llt x-a\,,\,y-2a\,,\,z-a^2\rgt =0 \qquad\text{or}\qquad 2ax -2ay +z = -a^2 \end{align*}

(b) The two planes are parallel when their two normal vectors, namely <2a , −2a , 1>\llt 2a\,,\,-2a\,,\,1\rgt and <1 , −1 , 1>\llt 1\,,\,-1\,,\,1\rgt, are parallel. This is the case if and only if a=12a=\frac{1}{2}.

Q10Stage 2Past exam · M200 2010D

Find the tangent plane and normal line to the surface z=f(x,y)=2yx2+y2z=f(x,y)=\frac{2y}{x^2+y^2} at (x,y)=(−1,2)(x,y)=(-1,2).

Answer

The tangent plane is 825x−625y−z=−85\frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5}.
The normal line is <x,y,z>=<−1,2,45>+t<825 , −625 , −1>\llt x,y,z\rgt = \llt -1,2,\frac{4}{5}\rgt +t \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt.

Full solution

The first order partial derivatives of ff are

fx(x,y)=−4xy(x2+y2)2fx(−1,2)=825fy(x,y)=2x2+y2−4y2(x2+y2)2fy(−1,2)=25−1625=−625\begin{alignat*}{3} f_x(x,y) & = -\frac{4xy}{{(x^2+y^2)}^2}\quad & f_x(-1,2) & = \frac{8}{25} \\ f_y(x,y) & = \frac{2}{x^2+y^2}-\frac{4y^2}{{(x^2+y^2)}^2}\quad & f_y(-1,2) & = \frac{2}{5}-\frac{16}{25} =-\frac{6}{25} \\ \end{alignat*}

So, by Theorem 2.5.1 in the CLP-3 text, a normal vector to the surface at (x,y)=(−1,2)(x,y)=(-1,2) is <825 , −625 , −1>\llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt. As f(−1,2)=45f(-1,2)= \frac{4}{5}, the tangent plane is

<825 , −625 , −1>⋅<x+1 , y−2 , z−45>=0or825x−625y−z=−85\begin{align*} \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt\cdot\llt x+1\,,\,y-2\,,\, z -\frac{4}{5}\rgt=0\quad \text{or}\quad \frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5} \end{align*}

and the normal line is

<x,y,z>=<−1,2,45>+t<825 , −625 , −1>\begin{align*} \llt x,y,z\rgt = \llt -1,2,\frac{4}{5}\rgt +t \llt \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1\rgt \end{align*}
Q11Stage 2Past exam · M200 2013D

Find all the points on the surface x2+9y2+4z2=17x^2 + 9y^2 + 4z^2 = 17 where the tangent plane is parallel to the plane x−8z=0x - 8z = 0.

Hint

Let (x,y,z)(x,y,z) be a desired point. Then

  • (x,y,z)(x,y,z) must be on the surface and

  • the normal vector to the surface at (x,y,z)(x,y,z) must be parallel to the plane's normal vector.

Answer

±(1,0,−2)\pm(1,0,-2)

Full solution

A normal vector to the surface x2+9y2+4z2=17x^2 + 9y^2 + 4z^2 = 17 at the point (x,y,z)(x,y,z) is <2x , 18y , 8z>\llt 2x\,,\, 18y\,,\,8z\rgt. A normal vector to the plane x−8z=0x - 8z = 0 is <1 , 0 , −8>\llt 1\,,\,0\,,\,-8\rgt. So we want <2x , 18y , 8z>\llt 2x\,,\, 18y\,,\,8z\rgt to be parallel to <1 , 0 , −8>\llt 1\,,\,0\,,\,-8\rgt, i.e. to be a nonzero constant times <1 , 0 , −8>\llt 1\,,\,0\,,\,-8\rgt. This is the case whenever y=0y=0 and z=−2xz=-2x with x≠0x\ne 0. In addition, we want (x,y,z)(x,y,z) to lie on the surface x2+9y2+4z2=17x^2 + 9y^2 + 4z^2 = 17. So we want y=0y=0, z=−2xz=-2x and

17=x2+9y2+4z2=x2+4(−2x)2=17x2  ⟹  x=±1\begin{align*} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{align*}

So the allowed points are ±(1,0,−2)\pm(1,0,-2).

Q12Stage 2Past exam · M200 2014D

Let SS be the surface z=x2+2y2+2y−1z = x^2 + 2y^2 + 2y - 1. Find all points P(x0,y0,z0)P (x_0,y_0,z_0) on SS with x0≠0x_0 \ne 0 such that the normal line at PP contains the origin (0,0,0)(0,0,0).

Hint

First find a parametric equation for the normal line to SS at (x0,y0,z0)(x_0,y_0,z_0). Then the requirement that (0,0,0)(0,0,0) lies on that normal line gives three equations in the four unknowns x0,y0,z0x_0,y_0,z_0 and tt. The requirement that (x0,y0,z0)(x_0,y_0,z_0) lies on SS gives a fourth equation. Solve this system of four equations.

Answer

(12 , −1 , −12)\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big) and (−12 , −1 , −12)\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)

Full solution

The equation of SS is of the form G(x,y,z)=x2+2y2+2y−z=1G(x,y,z) = x^2 + 2y^2 + 2y-z = 1. So one normal vector to SS at the point (x0,y0,z0)(x_0,y_0,z_0) is

∇G(x0,y0,z0)=2x0 ı^+(4y0+2) ȷ^−k^\begin{equation*} \vnabla G(x_0,y_0,z_0) = 2x_0\,\hi + (4y_0+2)\,\hj -\hk \end{equation*}

and the normal line to SS at (x0,y0,z0)(x_0,y_0,z_0) is

(x,y,z)=(x0,y0,z0)+t<2x0 , 4y0+2 , −1>\begin{equation*} (x,y,z) = (x_0,y_0,z_0) +t\llt 2x_0\,,\,4y_0+2\,,\, -1\rgt \end{equation*}

For this normal line to pass through the origin, there must be a tt with

(0,0,0)=(x0,y0,z0)+t<2x0 , 4y0+2 , −1>\begin{align*} (0,0,0) = (x_0,y_0,z_0) +t\llt 2x_0\,,\,4y_0+2\,,\, -1\rgt \end{align*}

or

x0+2x0 t=0y0+(4y0+2)t=0z0−t=0\begin{align*} x_0 + 2x_0\,t & =0 \tag{E1}\\ y_0 +(4y_0+2)t &=0 \tag{E2}\\ z_0 -t &=0 \tag{E3} \end{align*}

Equation (E3) forces t=z0t=z_0. Substituting this into equations (E1) and (E2) gives

x0(1+2z0)=0y0+(4y0+2)z0=0\begin{align*} x_0(1+2z_0) & =0 \tag{E1}\\ y_0 +(4y_0+2)z_0 &=0 \tag{E2} \end{align*}

The question specifies that x0≠0x_0\ne 0, so (E1) forces z0=−12z_0=-\frac{1}{2}. Substituting z0=−12z_0=-\frac{1}{2} into (E2) gives

−y0−1=0  ⟹  y0=−1\begin{equation*} -y_0-1=0 \implies y_0=-1 \end{equation*}

Finally x0x_0 is determined by the requirement that (x0,y0,z0)(x_0,y_0,z_0) must lie on SS and so must obey

z0=x02+2y02+2y0−1  ⟹  −12=x02+2(−1)2+2(−1)−1  ⟹  x02=12\begin{equation*} z_0 = x_0^2 + 2y_0^2 + 2y_0 - 1 \implies -\frac{1}{2} = x_0^2 + 2(-1)^2 +2(-1)-1 \implies x_0^2 = \frac{1}{2} \end{equation*}

So the allowed points PP are (12 , −1 , −12)\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big) and (−12 , −1 , −12)\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big).

Q13Stage 2Past exam · M226 2009D

Find all points on the hyperboloid z2=4x2+y2−1z^2=4x^2+y^2-1 where the tangent plane is parallel to the plane 2x−y+z=02x-y+z=0.

Hint

Two (nonzero) vectors v\vv and w\vw are parallel if and only if there
is a tt such that v=t w\vv=t\,\vw. Don't forget that the point has to be on the hyperboloid.

Answer

±(12,−1,−1)\pm \big(\half,-1,-1\big)

Full solution

Let (x0,y0,z0)(x_0,y_0,z_0) be a point on the hyperboloid z2=4x2+y2−1z^2=4x^2+y^2-1 where the tangent plane is parallel to the plane 2x−y+z=02x-y+z=0. A normal vector to the plane 2x−y+z=02x-y+z=0 is <2,−1,1>\llt 2,-1,1\rgt. Because the hyperboloid is G(x,y,z)=4x2+y2−z2−1G(x,y,z)=4x^2+y^2-z^2-1 and ∇G(x,y,z)=<8x,2y,−2z>\vnabla G(x,y,z) = \llt 8x,2y,-2z\rgt, a normal vector to the hyperboloid at (x0,y0,z0)(x_0,y_0,z_0) is ∇G(x0,y0,z0)=<8x0,2y0,−2z0>\vnabla G(x_0,y_0,z_0)=\llt 8x_0,2y_0,-2z_0\rgt. So (x0,y0,z0)(x_0,y_0,z_0) satisfies the required conditions if and only if there is a nonzero tt obeying

<8x0,2y0,−2z0>=t<2,−1,1> and z02=4x02+y02−1  ⟺  x0=t4, y0=z0=−t2 and z02=4x02+y02−1  ⟺  t24=t24+t24−1 and x0=t4, y0=z0=−t2  ⟺  t=±2(x0,y0,z0)=±(12,−1,−1)\begin{align*} &\llt 8x_0,2y_0,-2z_0\rgt =t\llt 2,-1,1\rgt \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\half,-1,-1\big) \end{align*}
Q14Stage 2

Find a vector of length 3\sqrt{3} which is tangent to the curve of intersection of the surfaces z2=4x2+9y2z^2=4x^2+9y^2 and 6x+3y+2z=56x+3y+2z=5 at (2,1,−5)(2,1,-5).

Hint

The curve lies in the surface z2=4x2+9y2z^2=4x^2+9y^2. So the tangent vector to the curve is perpendicular to the normal vector to z2=4x2+9y2z^2=4x^2+9y^2 at (2,1,−5)(2,1,-5).

The curve also lies in the surface 6x+3y+2z=56x+3y+2z=5. So the tangent vector to the curve is also perpendicular to the normal vector to 6x+3y+2z=56x+3y+2z=5 at (2,1,−5)(2,1,-5).

Answer

±3<3,14,−30>∣<3,14,−30>∣=±31105<3,14,−30>\pm\sqrt{3}\frac{\llt 3,14,-30\rgt}{|\llt 3,14,-30\rgt|} =\pm\sqrt{\frac{3}{1105}}\llt 3,14,-30\rgt

Full solution

One vector normal to the surface F(x,y,z)=4x2+9y2−z2=0F(x,y,z)=4x^2+9y^2-z^2=0 at (2,1,−5)(2,1,-5) is

∇F(2,1,−5)=<8x,18y,−2z>∣(2,1,−5)=<16,18,10>\begin{align*} \vnabla F(2,1,-5) = \llt 8x,18y,-2z\rgt\Big|_{(2,1,-5)}=\llt 16,18,10\rgt \end{align*}

One vector normal to the surface G(x,y,z)=6x+3y+2z=5G(x,y,z)=6x+3y+2z=5 at (2,1,−5)(2,1,-5) is

∇G(2,1,−5)=<6,3,2>\begin{align*} \vnabla G(2,1,-5) = \llt 6,3,2\rgt \end{align*}

Now

  • The curve lies in the surface z2=4x2+9y2z^2=4x^2+9y^2. So the tangent vector to the curve at (2,1,−5)(2,1,-5) is perpendicular to the normal vector 12<16,18,10>=<8,9,5>\frac{1}{2}\llt 16,18,10\rgt=\llt 8,9,5\rgt.

  • The curve also lies in the surface 6x+3y+2z=56x+3y+2z=5. So the tangent vector to the curve at (2,1,−5)(2,1,-5) is also perpendicular to the normal vector <6,3,2>\llt 6,3,2\rgt.

  • So the tangent vector to the curve at (2,1,−5)(2,1,-5) is parallel to

    <8,9,5>×<6,3,2>=det⁡[ı^ȷ^k^895632]=<3,14,−30>\begin{align*} \llt 8,9,5\rgt\times \llt 6,3,2\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 8 & 9 & 5 \\ 6 & 3 & 2 \end{matrix}\right] =\llt 3,14,-30\rgt \end{align*}

The desired vectors are

±3<3,14,−30>∣<3,14,−30>∣=±31105<3,14,−30>\begin{equation*} \pm\sqrt{3}\frac{\llt 3,14,-30\rgt}{|\llt 3,14,-30\rgt|} =\pm\sqrt{\frac{3}{1105}}\llt 3,14,-30\rgt \end{equation*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q15Stage 3

Find all horizontal planes that are tangent to the surface with equation

z=xye−(x2+y2)/2\begin{equation*} z=xy e^{-(x^2+y^2)/2} \end{equation*}

What are the largest and smallest values of zz on this surface?

Hint

At the highest and lowest points of the surface, the tangent plane is horizontal.

Answer

The horizontal tangent planes are z=0z=0, z=e−1z=e^{-1} and z=−e−1z=-e^{-1}. The largest and smallest values of zz are e−1e^{-1} and −e−1-e^{-1}, respectively.

Full solution

Let (x0,y0,z0)(x_0,y_0,z_0) be any point on the surface. A vector normal to the surface at (x0,y0,z0)(x_0,y_0,z_0) is

∇(xye−(x2+y2)/2−z)∣(x0,y0,z0)=<y0e−(x02+y02)/2−x02y0e−(x02+y02)/2,x0e−(x02+y02)/2−x0y02e−(x02+y02)/2,−1>\begin{align*} &\vnabla\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\&\hskip1in =\llt y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\rgt \end{align*}

The tangent plane to the surface at (x0,y0,z0)(x_0,y_0,z_0) is horizontal if and only if this vector is vertical, which is the case if and only if its xx- and yy-components are zero, which in turn is the case if and only if

y0(1−x02)=0 and x0(1−y02)=0  ⟺  {y0=0 or x0=1 or x0=−1} and {x0=0 or y0=1 or y0=−1}  ⟺  (x0,y0)=(0,0) or (1,1) or (1,−1) or (−1,1) or (−1,−1)\begin{align*} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{align*}

The values of z0z_0 at these points are 00, e−1e^{-1}, −e−1-e^{-1}, −e−1-e^{-1} and e−1e^{-1}, respectively. So the horizontal tangent planes are z=0z=0, z=e−1z=e^{-1} and z=−e−1z=-e^{-1}. At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of zz are e−1e^{-1} and −e−1-e^{-1}, respectively.

Q16Stage 3Past exam · M200 2004A

Let SS be the surface

xy−2x+yz+x2+y2+z3=7\begin{equation*} xy-2x+yz+x^2+y^2+z^3=7 \end{equation*}

  1. Find the tangent plane and normal line to the surface SS at the point (0,2,1)(0,2,1).

  2. The equation defining SS implicitly defines zz as a function of xx and yy for (x,y,z)(x,y,z) near (0,2,1)(0,2,1). Find expressions for ∂z∂x\pdiff{z}{x} and ∂z∂y\pdiff{z}{y}. Evaluate ∂z∂y\pdiff{z}{y} at (x,y,z)=(0,2,1)(x,y,z)=(0,2,1).

  3. Find an expression for ∂2z ∂x∂y\frac{\partial^2z\,}{\partial x\partial y}.

Answer

(a) y+z=3y+z=3 r(t)=<0,2,1>+t<0,5,5>\vr(t)=\llt 0,2,1\rgt+t\llt 0,5,5\rgt

(b) zx(x,y)=2−2x−yy+3z(x,y)2z_x(x,y)=\frac{2-2x-y}{y+3z(x,y)^2} zy(x,y)=−x+2y+z(x,y)y+3z(x,y)2z_y(x,y)=-\frac{x+2y+z(x,y)}{y+3z(x,y)^2} zy(0,2)=−1z_y(0,2)=-1

(c) zxy(x,y)=1y+3z(x,y)2−2−2x−y[y+3z(x,y)2]2(1−6z(x,y)x+2y+z(x,y)y+3z(x,y)2)z_{xy}(x,y)=\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2} \left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right)

Full solution

(a) A normal vector to the surface at (0,2,1)(0,2,1) is

∇(xy−2x+yz+x2+y2+z3−7)∣(0,2,1)=<y−2+2x , x+z+2y , y+3z2>∣(0,2,1)=<0,5,5>\begin{align*} \vnabla\big(xy-2x+yz+x^2+y^2+z^3-7\big)\big|_{(0,2,1)} &=\llt y-2+2x\,,\,x+z+2y\,,\,y+3z^2\rgt\big|_{(0,2,1)}\\ &=\llt 0, 5, 5\rgt \end{align*}

So the tangent plane is

0(x−0)+5(y−2)+5(z−1)=0 or y+z=3\begin{align*} 0(x-0)+5(y-2)+5(z-1)=0\text{ or }y+z=3 \end{align*}

The vector parametric equations for the normal line are

r(t)=<0,2,1>+t<0,5,5>\begin{equation*} \vr(t)=\llt 0,2,1\rgt+t\llt 0,5,5\rgt \end{equation*}

(b) Differentiating

xy−2x+y z(x,y)+x2+y2+z(x,y)3=7\begin{equation*} xy-2x+y\,z(x,y)+x^2+y^2+z(x,y)^3=7 \end{equation*}

gives

y−2+y zx(x,y)+2x+3z(x,y)2zx(x,y)=0  ⟹  zx(x,y)=2−2x−yy+3z(x,y)2x+z(x,y)+y zy(x,y)+2y+3z(x,y)2zy(x,y)=0  ⟹  zy(x,y)=−x+2y+z(x,y)y+3z(x,y)2\begin{alignat*}{5} y-2+y\,z_x(x,y)+2x+3z(x,y)^2z_x(x,y)&=0 & &\implies & z_x(x,y)&=\frac{2-2x-y}{y+3z(x,y)^2}\\ x+z(x,y)+y\,z_y(x,y)+2y+3z(x,y)^2z_y(x,y)&=0 & &\implies & z_y(x,y)&=-\frac{x+2y+z(x,y)}{y+3z(x,y)^2} \end{alignat*}

In particular, at (0,2,1)(0,2,1), zy(0,2)=−4+12+3=−1z_y(0,2)=-\frac{4+1}{2+3}=-1.

(c) Differentiating zxz_x with respect to yy gives

zxy(x,y)=−1y+3z(x,y)2−2−2x−y[y+3z(x,y)2]2(1+6z(x,y)zy(x,y))=−1y+3z(x,y)2−2−2x−y[y+3z(x,y)2]2(1−6z(x,y)x+2y+z(x,y)y+3z(x,y)2)\begin{align*} z_{xy}(x,y) &=-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2}\big(1+6z(x,y)z_y(x,y)\big)\\ &=-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2} \left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right) \end{align*}

As an alternate solution, we could also differentiate zyz_y with respect to xx. This gives

zyx(x,y)=−1+zx(x,y)y+3z(x,y)2+x+2y+z(x,y)[y+3z(x,y)2]26z(x,y)zx(x,y)=−1y+3z(x,y)2(1+2−2x−yy+3z(x,y)2)+x+2y+z(x,y)[y+3z(x,y)2]26z(x,y)2−2x−yy+3z(x,y)2\begin{align*} z_{yx}(x,y) &=-\frac{1+z_x(x,y)}{y+3z(x,y)^2} +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)z_x(x,y)\cr &=-\frac{1}{y+3z(x,y)^2}\left(1+\frac{2-2x-y}{y+3z(x,y)^2}\right) +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)\frac{2-2x-y}{y+3z(x,y)^2} \end{align*}
Q17Stage 3Past exam · M200 2000D

  1. Find a vector perpendicular at the point (1,1,3)(1,1,3) to the surface with equation x2+z2=10x^2+z^2=10.

  2. Find a vector tangent at the same point to the curve of intersection of the surface in part (a) with surface y2+z2=10y^2+z^2=10.

  3. Find parametric equations for the line tangent to that curve at that point.

Hint

(b) If v\vv is tangent, at a point PP, to the curve of intersection of the surfaces S1S_1 and S2S_2, then v\vv

  • has to be tangent to S1S_1 at PP, and so must be perpendicular to the normal vector to S1S_1 at PP and

  • has to be tangent to S2S_2 at PP, and so must be perpendicular to the normal vector to S2S_2 at PP.

Answer

(a) <1,0,3>\llt 1,0,3\rgt (b) <3,3,−1>\llt 3,3,-1\rgt (c) r(t)=<1,1,3>+t<3,3,−1>\vr(t)=\llt 1,1,3\rgt+t\llt 3,3,-1\rgt

Full solution

(a) A vector perpendicular to x2+z2=10x^2+z^2=10 at (1,1,3)(1,1,3) is

∇(x2+z2)∣(1,1,3)=(2xı^+2zk^)∣(1,1,3)=2ı^+6k^ or 12<2,0,6>=<1,0,3>\begin{equation*} \vnabla(x^2+z^2)\big|_{(1,1,3)} =(2x\hi+2z\hk)\big|_{(1,1,3)} =2\hi+6\hk\hbox{ or } \frac{1}{2} \llt 2,0,6\rgt=\llt 1,0,3\rgt \end{equation*}

(b) A vector perpendicular to y2+z2=10y^2+z^2=10 at (1,1,3)(1,1,3) is

∇(y2+z2)∣(1,1,3)=(2yȷ^+2zk^)∣(1,1,3)=2ȷ^+6k^ or 12<0,2,6>=<0,1,3>\begin{equation*} \vnabla(y^2+z^2)\big|_{(1,1,3)} =(2y\hj+2z\hk)\big|_{(1,1,3)} =2\hj+6\hk\hbox{ or }\frac{1}{2} \llt 0,2,6\rgt=\llt 0,1,3\rgt \end{equation*}

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both (1,0,3)(1,0,3) and (0,1,3)(0,1,3). One such vector is

<0,1,3>×<1,0,3>=det⁡[ı^ȷ^k^013103]=<3,3,−1>\begin{equation*} \llt 0,1,3\rgt\times\llt1,0,3\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =\llt 3,3,-1\rgt \end{equation*}

(c) The specified tangent line passes through (1,1,3)(1,1,3) and has direction vector <3,3,−1>\llt 3,3,-1\rgt and so has vector parametric equation

r(t)=<1,1,3>+t<3,3,−1>\begin{equation*} \vr(t)=\llt 1,1,3\rgt+t\llt 3,3,-1\rgt \end{equation*}
Q18Stage 3Past exam · M200 2000A

Let PP be the point where the curve

r(t)=t3 ı^+t ȷ^+t2 k^,(0≤t<∞)\begin{equation*} \vr(t) = t^3\,\hi + t\,\hj + t^2\,\hk,\qquad (0 \le t <\infty) \end{equation*}

intersects the surface

z3+xyz−2=0\begin{equation*} z^3 + xyz -2 = 0 \end{equation*}

Find the (acute) angle between the curve and the surface at PP.

Hint

The angle between the curve and the surface at PP is 90∘90^\circ minus the angle between the curve and the normal vector to the surface at PP.

Answer

49.11∘49.11^\circ (to two decimal places)

Full solution

r(t)=<x(t) , y(t) , z(t)>\vr(t)=\llt x(t)\,,\,y(t)\,,\,z(t)\rgt intersects z3+xyz−2=0z^3 + xyz -2 = 0 when

z(t)3+x(t) y(t) z(t)−2=0  ⟺  (t2)3+(t3)(t)(t2)−2=0  ⟺  2t6=2  ⟺  t=1\begin{equation*} z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1 \end{equation*}

since tt is required to be positive. The direction vector for the curve at t=1t=1 is

r′(1)=3 ı^+ȷ^+2 k^\begin{equation*} \vr'(1)=3\,\hi+\hj+2\,\hk \end{equation*}

A normal vector for the surface at r(1)=<1,1,1>\vr(1)=\llt 1,1,1\rgt is

∇(z3+xyz)∣(1,1,1)=[yzı^+xzȷ^+(3z2+xy)k^](1,1,1)=ı^+ȷ^+4k^\begin{equation*} \vnabla(z^3+xyz)\big|_{(1,1,1)}=[yz\hi+xz\hj+(3z^2+xy)\hk]_{(1,1,1)} =\hi+\hj+4\hk \end{equation*}

The angle θ\theta between the curve and the normal vector to the surface is determined by

∣<3,1,2>∣ ∣<1,1,4>∣cos⁡θ=<3,1,2>⋅<1,1,4>  ⟺  1418cos⁡θ=12  ⟺  7×36cos⁡θ=12  ⟺  cos⁡θ=27  ⟺  θ=40.89∘\begin{align*} \big|\llt 3,1,2\rgt\big|\,\big|\llt 1,1,4\rgt\big|\cos\theta =\llt 3,1,2\rgt \cdot\llt 1,1,4\rgt &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{align*}

The angle between the curve and the surface is 90−40.89=49.11∘90-40.89=49.11^\circ (to two decimal places).

Q19Stage 3

Find the distance from the point (1,1,0)(1,1,0) to the circular paraboloid with equation z=x2+y2z=x^2+y^2.

Hint

Let D(x,y)D(x,y) be the distance (or the square of the distance) from (1,1,0)(1,1,0) to the point (x,y,x2+y2)\big(x,y, x^2+y^2) on the paraboloid. We wish to minimize D(x,y)D(x,y). That is, to find the lowest point on the graph z=D(x,y)z=D(x,y). At this lowest point, the tangent plane to z=D(x,y)z=D(x,y) is horizontal.

Answer

32\frac{\sqrt{3}}{2}

Full solution

Let (x,y,z)(x,y,z) be any point on the paraboloid z=x2+y2z=x^2+y^2. The square of the distance from (1,1,0)(1,1,0) to this point is

D(x,y)=(x−1)2+(y−1)2+z2=(x−1)2+(y−1)2+(x2+y2)2\begin{align*} D(x,y)&=(x-1)^2+(y-1)^2+z^2\\ &=(x-1)^2+(y-1)^2+{(x^2+y^2)}^2 \end{align*}

We wish to minimize D(x,y)D(x,y). That is, to find the lowest point on the graph z=D(x,y)z=D(x,y). At this lowest point, the tangent plane to z=D(x,y)z=D(x,y) is horizontal. So at the minimum, the normal vector to z=D(x,y)z=D(x,y) has xx and yy components zero. So

0=∂D∂x(x,y)=2(x−1)+2(x2+y2)(2x)0=∂D∂y(x,y)=2(y−1)+2(x2+y2)(2y)\begin{alignat*}{5} 0&=\pdiff{D}{x}(x,y)&&= 2(x-1)+2(x^2+y^2)(2x)\\ 0&=\pdiff{D}{y}(x,y)&&= 2(y-1)+2(x^2+y^2)(2y) \end{alignat*}

By symmetry (or multiplying the first equation by yy, multiplying the second equation by xx and subtracting) the solution will have x=yx=y with

0=2(x−1)+2(x2+x2)(2x)=8x3+2x−2\begin{equation*} 0=2(x-1)+2(x^2+x^2)(2x)=8x^3+2x-2 \end{equation*}

Observe that the value of 8x3+2x−2=2(4x3+x−1)8x^3+2x-2=2(4x^3+x-1) at x=12x=\frac{1}{2} is 00. (See Appendix A.16 of the CLP-2 text for some useful tricks that can help you guess roots of polynomials with integer coefficients.) So (x−12)\big(x-\frac{1}{2}\big) is a factor of

4x3+x−1=4(x3+x4−14)=4(x−12)(x2+12x+12)\begin{equation*} 4x^3+x-1 ={\textstyle 4\big(x^3+\frac{x}{4}-\frac{1}{4}\big) =4\big(x-\frac{1}{2}\big)\big(x^2+\half x+\half\big)} \end{equation*}

and the minimizing (x,y)(x,y) obeys x=yx=y and

0=8x3+2x−2=8(x−12)(x2+12x+12)=0\begin{equation*} 0=8x^3+2x-2 =8\big(x-\half\big)\big(x^2+\half x+\half\big)=0 \end{equation*}

By the quadratic root formula, x2+12x+12x^2+\half x+\half has no real roots, so the only solution is x=y=12x=y=\half, z=(12)2+(12)2=12z=\big(\half\big)^2+\big(\half\big)^2=\half and the distance is (12−1)2+(12−1)2+(12)2=32\sqrt{\big(\half-1\big)^2+\big(\half-1\big)^2 +\big(\half\big)^2}=\frac{\sqrt{3}}{2}.

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.