Mathematics 241 – Multivariable Calculus Solutions for Exam 2 October 29, 2007

I. Let F (x, y) = (x − y^{2}, xy) and g(t) = (sin(t), cos(t) + sin(t)). A) Compute the derivative matrices D(F) and D(g).

Solution: By the definition, D(F) is a 2 × 2 matrix and D(g) is a 2 × 1 matrix:

D(F) = ( \begin{matrix}\frac{∂}{∂x}(x − y^{2}) & \frac{∂}{∂y} (x − y^{2}) \\ \frac{∂}{∂x}(xy) & \frac{∂}{∂y} (xy) \end{matrix} ) = ( \begin{matrix}1 & −2y \\ y & x \end{matrix} )

and

D(g) = ( \begin{matrix}cos(t) \\ − sin(t) + cos(t) \end{matrix} ) .

B) Use the Chain Rule to compute D(F ∘ g)(π).

Solution: The Chain Rule says D(F ∘ g)(π) = D(F)(g(π))D(g)(π) here. We have g(π) = (0, −1) so

D(F)(g(π)) = ( \begin{matrix}1 & 2 \\ −1 & 0 \end{matrix} )

and

D(g)(π) = ( \begin{matrix}−1 \\ −1 \end{matrix} ) .

So

D(F)(g(π))D(g)(π) = ( \begin{matrix}1 & 2 \\ −1 & 0 \end{matrix} ) ( \begin{matrix}−1 \\ −1 \end{matrix} ) = ( \begin{matrix}−3 \\ 1 \end{matrix} ) .

C) Compute the composition F ∘ g and find its derivative matrix directly to check your answer from part B.

Solution: We have

(F ∘ g)(t) = (sin(t) − (cos(t) + sin(t))^{2}, sin(t)(cos(t) + sin(t)).

So

D(F ∘ g)(t) = ( \begin{matrix}cos(t) − 2(cos(t) + sin(t))(− sin(t) + cos(t)) \\ sin(t)(− sin(t) + cos(t)) + (cos(t) + sin(t)) cos(t) \end{matrix} )

Substituting t = π, we get:

D(F ∘ g)(π) = ( \begin{matrix}cos(π) − 2(cos(π) + sin(π))(− sin(π) + cos(π)) \\ sin(π)(− sin(π) + cos(π)) + (cos(π) + sin(π)) cos(π) \end{matrix} )

= ( \begin{matrix}−1 − 2(−1)(−1) \\ 0 + (−1)(−1) \end{matrix} )

= ( \begin{matrix}−3 \\ 1 \end{matrix} ) ,

since sin(π) = 0 and cos(π) = −1.

II. Let z = f(x + y, x − y) have continuous partial derivatives with respect to u = x + y and v = x − y. Show that

\frac{∂z}{∂x} \frac{∂z}{∂y} = (\frac{∂z}{∂u})^{2} − (\frac{∂z}{∂v} )^{2} .

Solution: By the Chain Rule,

\frac{∂z}{∂x} = \frac{∂z}{∂u} \frac{∂u}{∂x} + \frac{∂z}{∂v} \frac{∂v}{∂x} = \frac{∂z}{∂u} · 1 + \frac{∂z}{∂v} · 1

and

\frac{∂z}{∂y} = \frac{∂z}{∂u} \frac{∂u}{∂y} + \frac{∂z}{∂v} \frac{∂v}{∂y} = \frac{∂z}{∂u} · 1 + \frac{∂z}{∂v} · (−1).

Then

\frac{∂z}{∂x} \frac{∂z}{∂y} = (\frac{∂z}{∂u} + \frac{∂z}{∂v} ) (\frac{∂z}{∂u} − \frac{∂z}{∂v} ) = (\frac{∂z}{∂u})^{2} − (\frac{∂z}{∂v} )^{2} .

This establishes what we wanted to show.

Comment: It is also possible to use the Chain Rule a different way to do this problem. If solve the equations u = x + y and v = x − y for x, y in terms of u, v, then we find x = \frac{1}{2}(u + v) and y = \frac{1}{2}(u − v). Then by the Chain Rule,

\frac{∂z}{∂u} = \frac{∂z}{∂x} \frac{∂x}{∂u} + \frac{∂z}{∂y} \frac{∂y}{∂u} = \frac{∂z}{∂x} · \frac{1}{2} + \frac{∂z}{∂y} · \frac{1}{2}

and

\frac{∂z}{∂v} = \frac{∂z}{∂x} \frac{∂x}{∂v} + \frac{∂z}{∂y} \frac{∂y}{∂v} = \frac{∂z}{∂x} · \frac{1}{2} + \frac{∂z}{∂y} · \frac{−1}{2}.

Then substituting and expanding the squares

(\frac{∂z}{∂u})^{2} − (\frac{∂z}{∂v} )^{2} = (\frac{∂z}{∂x} · \frac{1}{2} + \frac{∂z}{∂y} · \frac{1}{2})^{2} − (\frac{∂z}{∂x} · \frac{1}{2} + \frac{∂z}{∂y} · \frac{−1}{2})^{2}

= \frac{1}{4} (\frac{∂z}{∂x})^{2} + \frac{1}{2} \frac{∂z}{∂x} \frac{∂z}{∂y} + \frac{1}{4} (\frac{∂z}{∂y} )^{2}

− (\frac{1}{4} (\frac{∂z}{∂x})^{2} − \frac{1}{2} \frac{∂z}{∂x} \frac{∂z}{∂y} + \frac{1}{4} (\frac{∂z}{∂y} )^{2})

= \frac{∂z}{∂x} \frac{∂z}{∂y} .

This establishes what we wanted to show again, but note the first way is somewhat simpler.

III. A mountain has the shape of the graph

z = \sqrt{4 − x^{2} − \frac{y^{2}}{4}} ,

where x, y are the east-west and north-south map coordinates respectively, and z is the altitude. A) If you are standing at the point on the mountain with (x, y, z) = (1, 2, \sqrt{2}), which

direction should you move (on the map) to increase your altitude the fastest?

Solution: This will be in the direction of the gradient vector of the altitude function at (x, y) = (1, 2). We have

\begin{aligned}∇f(x, y) & = (\frac{∂}{∂x}\sqrt{4 − x^{2} − \frac{y^{2}}{4}} , \frac{∂}{∂y} \sqrt{4 − x^{2} − \frac{y^{2}}{4}} ) \\ & = (\frac{−x}{\sqrt{4 − x^{2} − \frac{y^{2}}{4}}} , \frac{−y}{4\sqrt{4 − x^{2} − \frac{y^{2}}{4}}} ) , \end{aligned}

so at (x, y) = (1, 2), we have

∇f(1, 2) = (\frac{−1}{\sqrt{2}}, \frac{−1}{2\sqrt{2}}) .

The unit vector specifying the direction of the gradient is

u = (\frac{−2}{\sqrt{5}}, \frac{−1}{\sqrt{5}}) .

B) At the same point as in part A, in what direction should you move (on the map) to remain at a constant altitude?

Solution: This will be in directions orthogonal to the gradient vector. These are specified by the unit vectors

u = ± (\frac{1}{\sqrt{5}}, \frac{−2}{\sqrt{5}}) .

(Note: Given (a, b), to get a perpendicular vector (c, d) with (a, b) · (c, d) = 0, we can always take (c, d) = (−b, a).)

IV. Let x(t) = (t^{2}/2, t, t) and y(t) = (t cos(t), t sin(t), t) be two curves in R^{3}. A) Does the tangent line to x(t) at t = 2 cross the xy-plane? If so, where does this occur?

Solution: The tangent line is the line with parametric equations

x(2) + (t − 2)x′(2) = (2, 2, 2) + (t − 2)(2, 1, 1) = (2t − 2, t, t)

This crosses the plane z = 0 at t = 0, and the intersection point is (−2, 0, 0).

B) Show that y(t) (for t > 0) is a flow line for the vector field

F (x, y, z) = (−y + \frac{x}{z}, x + \frac{y}{z} , 1) .

Solution: The curve α(t) is a flow line of F if α′(t) = F (α(t)). We have y′(t) = (−t sin(t) + cos(t), t cos(t) + sin(t), 1). This is the same as what we get by substituting the component functions of the curve into the formula for F: (−t sin(t) + cos(t), t cos(t) + sin(t), 1).

C) Show that the segments of x(t) and y(t) for a ≤ t ≤ b have the same arclength for all a, b.

Solution: We have the arclength for x(t) is

∫_{a}^{b} ∥x′(t)∥ dt = ∫_{a}^{b} \sqrt{t^{2} + 2} dt.

Similarly for y(t),

∥y′(t)∥ = \sqrt{(−t sin(t) + cos(t))^{2} + (t cos(t) + sin(t))^{2} + 1}
= \sqrt{(t^{2} + 1)(sin^{2}(t) + cos^{2}(t)) − 2t sin(t) cos(t) + 2t sin(t) cos(t) + 1}
= \sqrt{t^{2} + 2}.

Hence the arclength for y(t) is ∫_{a}^{b} \sqrt{t^{2} + 2} dt as well.

Extra Credit: The arclength function for x(t) is

∥dT/dt∥/∥dx/dt∥ = \frac{\sqrt{2}}{(t^{2} + 2)^{3/2}}

(after simplification!)