SOLUTIONS TO PRACTICE EXAM 2

1. Find the directional derivative of f(x, y) = x^{3}y+xy^{2} at the point (1, 2) in the direction of 3i⃗ − 4j⃗. f_{x} = 3x^{2}y + y^{2} and f_{y} = x^{3} + 2xy, so that gradf(1, 2) = ⟨10, 5⟩. Furthermore

normalizing v⃗ = 3i⃗ − 4j⃗ we obtain u⃗ = ⟨3/5, −4/5⟩, so that fv⃗(1, 2) =grad f(1, 2) · u⃗ = ⟨10, 5⟩ · ⟨3/5, −4/5⟩ = 6 − 4 = 2.

2. Suppose a heat-seeking bug is put down at the origin on the xy-plane where the temperature is given by T (x, y) = 50 + 2x − x^{2} − y^{2} in ^{∘}C.

a) If the bug starts moving in the direction making an angle \frac{π}{4} with the x-axis, what is the rate of change of temperature in this direction?

T_{x} = 2 − 2x, T_{y} = −2y and u⃗ = \frac{1}{\sqrt{2}}⟨1, 1⟩, so that Tu⃗(0, 0) = T_{x}(0, 0)u_{1} + T_{y}(0, 0)u_{2} = 2/\sqrt{2} + 0.

b) In which direction should the bug go to ensure the fastest increase of temperature? ∇T (0, 0) = ⟨2, 0⟩, so that moving in positive x-direction ensures the greatest increase

in temperature.

c) Where on the plane is it hottest? Setting ∇T = 0, we get that (x, y) = (1, 0) is a critical point. D = (−2)(−2) + 0^{2} > 0,

and T_{xx} < 0, so this indeed a local maximum. To see that it is also a global maximum it suffices to notice that the surface looks like a bowl opening downwards.

3. True or false:

a) There is a function with ∥grad f∥ = 4 and fi⃗ = 5 at some point. False, the greatest slope is in the direction of the gradient and is ∥grad f∥ = 4, but

fi⃗ = 5 would be bigger then that.

b) There is a function with ∥grad f∥ = 4 and fj⃗ = −3 at some point. That is true, f(x, y) = x\sqrt{7} − 3y has that property at the origin.

c) If grad f = grad g then f = g. False, see quiz 3.

4. Compute \frac{∂^{2}f}{∂x∂y} if f(x, y) = (xy)^{7}.

f_{y}(x, y) = 7(xy)^{6}x and so f_{yx}(x, y) = 42(xy)^{5}y(x) + 7(xy)^{6}(1).

5. Let Q(x, y) = A + Bx + Cy + Dx^{2} + Exy + F y^{2} be the quadratic approximation to the function f(x, y) near the origin. Express \frac{∂^{2}Q}{∂x∂y} in terms of derivatives of f.

\frac{∂^{2}Q}{∂x∂y} = E = f_{xy}(0, 0).

6. Suppose f(x, y) = xe^{y}. We can consider it as a function of polar coordinates r and θ via x = r cos θ and y = r sin θ. Find \frac{∂f}{∂r} using the chain rule.

\frac{∂f}{∂r} = \frac{∂f}{∂x} \frac{∂x}{∂r} + \frac{∂f}{∂y} \frac{∂y}{∂r} = e^{y} cos θ + xe^{y} sin θ = e^{r sin θ} cos θ + r cos θe^{r sin θ} sin θ.

7. Is (0, 0) a critical point of the following functions:

a) f(x, y) = cos x sin y If “yes”: min max or saddle?

No, since f_{y}(0, 0) = cos 0 cos 0 = 1 ≠ 0.

b) f(x, y) = sin x sin y If “yes”: min max or saddle?

Yes, f_{y}(x, y) = sin x cos y and f_{x}(x, y) = cos x sin y, so that f_{x}(0, 0) = 0 and f_{y}(0, 0) = 0. It is a saddle point, since D = (− sin x sin y)(− sin x sin y) − (cos x cos y)^{2} = −1 at (0,0).

8. Find the (minimal) distance between the point (1, 2, 3) and the plane z = 5x − y + 1.

Either argue using the normal vector of the plane, or minimize the square of the distance between (1, 2, 3) and (x, y, z) on the plane:

(x − 1)^{2} + (y − 2)^{2} + (z − 3)^{2} = (x − 1)^{2} + (y − 2)^{2} + (5x − y + 1 − 3)^{2} = f(x, y).

f_{x}(x, y) = 2(x − 1) + 2(5x − y − 2)5 = 0 and f_{y}(x, y) = 2(y − 2) + 2(5x − y − 2)(−1) = 0 yield that (22/27, 55/27) is the only critical point and D = 52 · 4 − 10^{2} > 0, yields that it is indeed a local minimum, and by the nature of the situation therefore also the global minimum. Since z = 5x−y +1 = 82/27, we get that the minimum distance is the distance between these two points, which is 1/\sqrt{27}.