# Partial Solutions to Brainfood #3

These are partial answers so that you can check to see if you are on target. We expect your actual answers to be "fleshed out" a bit more.
1. When w=2, a unit change in length results in a change of 2 for the area function.
2. Along the diagonal, each value in the table corresponds to a change of sqrt(2) in the domain. Moving from (1,1) to (2,2) results in a change of 3, whereas moving from (2,2) to (3,3) results in a change of 5. Averaging these changes, we we estimate the directional derivative at (2,2) in the direction of the diagonal to be 4/sqrt(2).
3. d/dl of A(l,2) is 2 for all values of l. d/dl of A(l,l) is 2l, so at (2,2) the instantaneous rate of change in the direction of the diagonal is 4, whereas at (4,4) the rate of change is 8.
5. It's okay if you were stumped on this question; it looks ahead to the multivariable chain rule (which we have not yet discussed, see p. 115). This problem is similar to the example on p. 183.

The key to this problem is that the words "expected change in a function for a unit change in a variable at variable=X" MEANS "take the derivative of the function with respect to the variable and evaluate at variable=X". It DOES NOT mean "evaluate the function at X-1 and X+1 and use those values to estimate the change at X".

The main idea is that if you restrict yourself to a parametrized curve, C, then the rate of change of a function restricted to C depends on how fast we move along C. Think about moving slowly along a roller-coaster track versus moving quickly over the same track: the faster you move, the "bigger" the drops and hills seem, even though the hills do not change their size.

A unit change in t makes a change of (l'(t), w'(t)) along the parametrized curve. This tangent vector determines the direction that we will use for the directional derivative. At t=3 Pi/2, the tangent vector to the circle is T=(2,0). Note that the norm of T is 2. But recall that the directional derivative (which we computed in question 1) is the change for a UNIT step in some direction. Since we are taking a step of length 2, the expected change is the area is TWICE the directional derivative in the direction (1,0).

Here's a summary of the answer:

• Note that C(3*Pi/2) = (2,2).
• For a unit change in t, the expected change in (l,w)-space is the derivative evaluated at the time of interest: C'(3 Pi/2) = (2,0).
• If we restrict A to C, then a unit change in t will lead to an expected change of dA/dt evaluated at t=3*Pi/2. We can symbolically restrict A to C by just substituting the parametrization of C into A to get A(l(t),w(t))=(2+2cos(t))(4+2sin(t)). This is just a function of t, so take the derivative and evaluate at t=3*Pi/2 to obtain 4 as the instantaneous rate of change.
• Note that in question 1 we saw that the expected change for a UNIT change in l was 2. But as we've seen in part b, a change in t leads to a change by TWO UNITS in l. Therefore a unit change in t leads to twice the change that we saw in problem 1.
Here's a sketch of the graph of A restricted to C.