18.02 Practice Exam 2 -- March, 1997 partial solutions soon

Problem 1 (30 points) All six parts refer to the function

$$f(x,y)= \frac{y}{x} \, .$$

1.

Draw five reasonably spaced level curves for $f(x,y)$ in the $xy$-plane; label each with the corresponding value of $f(x,y)$.

2.

Let $w=f(x,y)$. Find $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$.

3.

Find the gradient vector $\overrightarrow{\triangledown w}$ at any point $(a,b)$ where it is defined, and show by calculation that it is perpendicular to the level curve of $f(x,y)$ passing through $(a,b)$.

4.

Find the directional derivative $\frac{dw}{ds}$ at the point $(2,1)$ in the direction of the vector $3{\bf i}+4{\bf j}$.

If you start at this point and go in this direction, approximately what distance would you travel to decrease the value of $w$ by $.02$?

5.

Find the point $P$ on the graph of $f(x,y)$ lying over the point $(2,1)$ in the $xy$-plane, and find the tangent plane to the graph at the point $P$.

6.

Give an approximate expression for $\Delta w$ in terms of $\Delta x$ and $\Delta y$, for values of $(x,y)$ close to $(2,1)$. Use it to answer the two questions below.

Near $(2,1)$, is the value of $w$ more sensitive to $x$ or to $y$?

Near $(2,1)$, if an error of $\pm .01$ is made measuring $x$ and $y$, what is the possible resulting error in the corresponding value of $w$?

Partial solution: b)

$$\frac{\partial w}{\partial x}=-\frac y{x^2},\ \frac{\partial w}{\partial y}=\frac1x c)$$

$$\overrightarrow{\triangledown w}=-\frac b{a^2}{\bf i}+\frac1a{\bf j},\ b\not=0$$

The level curve through $(a,b)$ is $y/x=b/a$ that is $x=at,$ $y=bt$ which is a straightline with tangent vector $a{\bf i}+b{\bf j}.$

$$\overrightarrow{\triangledown w}\cdot(a{\bf i}+{\bf j})=-\frac ba+\frac ba=0.$$

d) The directional derivative is $dw/ds=\overrightarrow{\triangledown w}\cdot\hat t$ where $\hat t$ is a unit vector in give direction. Thus the directional derivative is

$$\frac{dw}{ds}={-\frac14{\bf i}+\frac12{\bf j}}\cdot\frac{3{\bf i}+4{\bf j}}{5}=1/4.$$

Since $\Delta w\sim\frac{dw}{ds}\Delta s,$ should go approximately $.08$ backwards.

e) Point on graph is $P=(2,1,1/2).$ Tangent plane to $z-y/x=0$ is

$$(z-\frac12)+\frac14(x-2)-\frac12(y-1)=0.$$

f)

$$\Delta w\sim -\frac14\Delta x+\frac12\Delta y$$

is more sensitive to changes in $y.$ Possible error is

$$-\frac14\times(-0.01)+\frac12\times0.01=.008.$$

Problem 2 (30 points) A box is to be constructed from wood pieces of uniform thickness so that the top and two opposite sides use a single thickness of wood, while the two ends and the bottom use a double thickness. The volume is to be $24$ cubic feet.

What dimensions for the box will use the least amount of wood?

1.

Using $z$ for the height and $x,y$ for the other two dimensions, show the function to be minimized has the form ($x$ and $y$ could be interchanged):

$$w=3xy+96/x+48/y \, .$$

2.

Find the $x$, $y$ and $z$ values which minimize $w$.

3.

If the problem is solved using Lagrange multipliers, what is the value of the multiplier $\lambda$ corresponding to the minimum? (To solve this, you do not need to write down all the equations required by the method.)

Partial solution: a) The amount of wood is

$$w=xy+2xy+4xz+2yz=3xy+\frac{96}y+\frac{48}x$$

from top + bottom + sides + ends and given that $xyz=24.$

b) $\partial w/\partial x=3y-48/x^2=0$ and $\partial w/\partial y=3x-96/y^2=0.$ Solving gives $x=2,$ $y=4$ and $z=3.$

c) One Lagrange equation is $\partial w/\partial x=3y+4x=\lambda yz$ so $\lambda=2.$

Problem 3 (10 points) Suppose a change from $xy$-coordinates to $uv$-coordinates is given by the equations

$$x=u^2-v^2 , \qquad y=2uv \, .$$

Let $w=f(x,y)$. Then after the change of coordinates, $w$ becomes a function of $u$ and $v$ which we shall denote by $g(u,v)$, that is,

$$w=f(x,y) = f(u^2 - v^2, 2uv) = g(u,v) \, .$$

1.

Using the chain rule, express $\frac{\partial g}{\partial u}$ and $\frac{\partial g}{\partial v}$ in terms of $u,v, \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}$.

2.

If $\overrightarrow{\triangledown f} = (3,1)$ at the point $(0,2)$ in the $xy$-plane, what is $\overrightarrow{\triangledown g}$ at the point $(1,1)$ in the $uv$-plane?

Problem 4 (10 points) Where does the tangent plane to the surface $x^2 + 3y^2 + 4z^2 =8$ at the point $(1,1,1)$ intersect the $z$-axis?

Partial solution: Tangent plane is $2(x-1)+6(y-1)+8(z-1)=0$ intersects the z-axis at $x=y=0$ that is $z=2.$

Problem 5 (10 points) Suppose $w=f(x,y,z)$, where $y^2=xz$. Express $\left( \frac{\partial w}{\partial x} \right)_y$ in terms of the partial derivatives $f_x$, $f_y$, and $f_z$.

Partial solution: Since $y^2=xz,$ $z=y^2/x$ so $\left(\frac{\partial z}{\partial x}\right)_y=-\frac{y^2}{x^2}.$

$$\left(\frac{\partial w}{\partial x}\right)_y=f_x+f_z\left(\frac{\partial z}{\partial x}\right)_y = f_x+z_z\times(-\frac{y^2}{x^2}).$$

Problem 6 (10 points) Let $(f(x,y)$ denote the height of the point $(x,y)$ above sea level. A hiker is ascending a hill. The motion of the hiker is (as observed in the $xy$-plane, i.e., on a topographic map of the hill) has these two properties:

1.

It is always in the direction of $\overrightarrow{\triangledown f}$, i.e., perpendicular to the level curves of $f(x,y)$;

2.

Its speed (in the $xy$-plane is inversely proportional to $\left\vert \overrightarrow{\triangledown f} \right\vert$.

Show that the hiker is ascending at a constant rate.

(Express the velocity vector in terms of $f$, then use the chain rule to calculate $df/\, dt$.)