MATH 285, HONORS MULTIVARIABLE CALCULUS, FALL 1999 Answers to the Second Midterm

Problem 1 (8 points) Use a tangent plane approximation to approximate (0.99)^{3}e^{0.01}.

Answer: Let f(x, y) = x^{3}e^{y}. The tangent plane approximation for f near a point (x_{0}, y_{0}) is

f(x, y) ≈ f(x_{0}, y_{0}) + f_{x}(x_{0}, y_{0}) Δx + f_{y}(x_{0}, y_{0}) Δy,

where Δx = x − x_{0} and Δy = y − y_{0}. We calculate f_{x}(x, y) = 3x^{2}e^{y} and f_{y}(x, y) = x^{3}e^{y}. If we take (x_{0}, y_{0}) = (1, 0), we get:

f(0.99, 0.01) ≈ f(1, 0) + f_{x}(1, 0) · (−0.01) + f_{y}(1, 0) · 0.01
= 1 − 3 · 0.01 + 1 · 0.01
= 0.98.

Problem 2 (8 points) Find the average value of the function f(x) = ∫_{x/2}^{1} sin(t^{2}) dt on the interval [0, 2].

Answer: The average value of f(x) on the interval [0, 2] is

\frac{1}{2} ∫_{0}^{2} f(x) dx = \frac{1}{2} ∫_{0}^{2} ∫_{x/2}^{1} sin(t^{2}) dt dx
= \frac{1}{2} ∫_{0}^{1} ∫_{0}^{2t} sin(t^{2}) dx dt
= \frac{1}{2} ∫_{0}^{1} 2t sin(t^{2}) dt
= \frac{1}{2} [− cos(t^{2})]_{0}^{1}
= \frac{1}{2} − \frac{1}{2} cos 1.

Problem 3 Suppose f(x, y) is a function such that ∇f = ⟨x^{2} − y, −x − y + 2⟩.

(a) (5 points) Use the nullcline method to sketch the vector field ∇f.

Answer:

(b) (5 points) Determine all local maxima, minima, and saddle points of f. Does f have a global maximum or minimum?

Answer: The critical points of f occur where f_{x} = 0 and f_{y} = 0. In the above sketch, we see that this happens at (x, y) = (−2, 4) and (x, y) = (1, 1). We do the second derivative test: D = f_{xx}f_{yy}−f_{xy}^{2} = 2x·(−1)−(−1)^{2} = 1−2x. We find that D is negative at (x, y) = (1, 1), making (1, 1) a saddle point. Also, D is positive at (x, y) = (−2, 4), making (−2, 4) into a local maximum or minimum. Since f_{x}x(−2, 4) = −4 is negative, we find that f(−2, 4) is a local maximum. This can also be seen by looking at the above sketch and using the fact that the gradient always points in the direction of greatest increase.

f has no global maxima or minima. If we keep y = 0 fixed, we find that df/dx = x^{2}, thus f(x, 0) = x^{3}/3 + C, thus lim_{x→∞} f(x, 0) = +∞ and lim_{x→−∞} f(x, 0) = −∞.

Problem 4 (10 points) Franny says: “I know a differentiable function f(x, y) such that

\frac{∂}{∂x}f(x, y) = xe^{x^{2}y^{2}}  and  \frac{∂}{∂y} f(x, y) = ye^{x^{2}y^{2}}.

Zooey says: “And I know a differentiable function g(x, y) such that

\frac{∂}{∂x}g(x, y) = ye^{x^{2}y^{2}}  and  \frac{∂}{∂y} g(x, y) = xe^{x^{2}y^{2}}.

One of them is lying. Which one, and how do you know?

Answer: One way to check this is by Clairaut’s Theorem. Checking Franny’s function, we have

\frac{∂}{∂y} \frac{∂}{∂x}f(x, y) = xe^{x^{2}y^{2}} · 2yx^{2},  and  \frac{∂}{∂x} \frac{∂}{∂y} f(x, y) = ye^{x^{2}y^{2}} · 2xy^{2}.

If Franny really had such a function f(x, y), then by Clairaut’s Theorem, the two expressions would have to be equal, but they are not. So Franny is lying. On the other hand, for Zooey’s function, we get

\frac{∂}{∂y} \frac{∂}{∂x}g(x, y) = e^{x^{2}y^{2}} + ye^{x^{2}y^{2}} · 2yx^{2},  and  \frac{∂}{∂x} \frac{∂}{∂y} g(x, y) = e^{x^{2}y^{2}} + xe^{x^{2}y^{2}} · 2xy^{2}.

These functions are indeed equal, which makes Zooey’s claim plausible. You did not have to find the function g, but in case you’re curious, here is a possible definition:

g(x, y) = ∫_{0}^{xy} e^{t^{2}}dt.

Problem 5 Consider the function f which is defined as follows:

f(x, y) = \{ \begin{matrix}1 & if x ≠ 0 and y = x^{2} \\ 0 & otherwise. \end{matrix}

(a) (4 points) Is f differentiable at (0, 0)?

Answer: f is not continuous at (0, 0), because lim_{x→0} f(x, x^{2}) = lim_{x→0} 1 = 1, whereas f(0, 0) = 0. Since f is not continuous, it is certainly not differentiable.

(b) (10 points) Prove that for every unit vector u⃗, the directional derivative Du⃗f(0, 0) exists.

Answer: Informally, the reason is that any fixed line through the origin has a small stretch near (0,0) where f is constantly zero.

More formally: Let u⃗ = ⟨a, b⟩ be any fixed unit vector. The directional derivative Du⃗f(0, 0) is defined as the limit

lim_{h→0} \frac{f(ha, hb) − f(0, 0)}{h} = lim_{h→0} \frac{f(ha, hb)}{h}.

We consider two cases: Case 1: a = 0. In this case, f(ha, hb) = f(0, hb) = 0 for all h, thus, the above limit exists and is equal to zero. Case 2: a ≠ 0. By definition of f, we have f(ha, hb) = 1 only when hb = (ha)^{2} and ha ≠ 0, i.e., when hb = h^{2}a^{2}, which happens only at h = \frac{b}{a^{2}} . Thus, when h is sufficiently small (namely, when |h| < |\frac{b}{a^{2}} |), then we have f(ha, hb) = 0. It follows that lim_{h→0} \frac{f(ha,hb)}{h} = 0.

In either case, the limit exists.