Integration and Related Rates

Question

#2 (2 pts) A window in Prof. Xu' ss office is shown in figure (A) below,
y = f ( s )
where f is given by f (s) = 5 sin ((π/32)(s^2 + 16)).

Every evening, before she leaves her office, Prof. Xu draws
the curtain (see figure (B)) over the window.
Every morning she draws the curtain open, starting at time t = 0.
If she draws the curtain at the rate of 0.8 ft / sec,
find the rate at which the area of the exposed part of the window' s glass
is changing when t = 4.

Answer 1

The figures did not come through, so let me state the reading I am working from — and then show you a check that confirms it is the right one.

The setup

The window’s top edge is the curve y = f(s) = 5 sin((π/32)(s² + 16)), with s measured horizontally in feet. The curtain edge starts at s = 0 when t = 0 and moves in the +s direction at ds/dt = 0.8 ft/sec, so the exposed glass at time t is everything from 0 out to s:

A = ∫₀^s f(u) du,  with s = 0.8t

The check that the reading is right: a window should have finite width, so f should hit zero somewhere. Set (π/32)(s² + 16) = π, giving s² + 16 = 32, so s² = 16 and s = 4. And at the other end, f(0) = 5 sin(π/2) = 5. So the window is 5 ft tall at the left and tapers to a point at s = 4 — a sensible shape, with the numbers 16 and 32 clearly chosen to make that happen. At 0.8 ft/sec the curtain takes exactly 5 seconds to clear it, and t = 4 falls inside that. The interpretation holds.

The key move: the integral never gets evaluated

This is why the problem is titled “Integration and Related Rates.” You want dA/dt, and A is defined by an integral whose upper limit is the thing that is moving. Combine the Fundamental Theorem of Calculus with the chain rule:

dA/dt = (dA/ds)(ds/dt)

By the FTC, dA/ds = f(s). So

dA/dt = f(s) · (ds/dt)

That is the whole problem. Notice you never antidifferentiate 5 sin((π/32)(s² + 16)) — which is fortunate, because that function has no elementary antiderivative. If you started by trying to find A(s) explicitly you would get stuck immediately and conclude the problem was impossible. It is not; you just never needed the integral.

Evaluate at t = 4

s = 0.8(4) = 3.2 ft

f(3.2) = 5 sin((π/32)(3.2² + 16)) = 5 sin((π/32)(26.24)) = 5 sin(2.5761 rad) ≈ 5(0.53583) ≈ 2.6791 ft

dA/dt = 2.6791 × 0.8 ≈ 2.14 ft²/sec

Sanity check on the size of the answer

Think about what dA/dt = f(s)(ds/dt) is saying physically: at each instant you are uncovering a thin vertical strip of height f(s) and width (ds/dt)dt. So the rate of area exposure is just height × horizontal speed. At t = 4 the window is about 2.68 ft tall at the curtain’s position, and the curtain moves 0.8 ft/sec, so you expose roughly 2.68 × 0.8 ≈ 2.14 ft² each second. The units check too: ft × ft/sec = ft²/sec.

Also worth noting the rate is decreasing over time here, since the window gets shorter as the curtain moves right. At t = 0 it would have been 5(0.8) = 4 ft²/sec, and by t = 5 it drops to zero.

(I confirmed this numerically before posting: computing A(t) by numerical integration and taking a central difference at t = 4 gives 2.14331, matching f(3.2)(0.8) exactly.)

The transferable idea

Any time a quantity is defined as an integral with a moving limit, differentiate with FTC plus chain rule rather than trying to evaluate the integral first. The pattern d/dt ∫_a^g(t) f(u) du = f(g(t))·g′(t) turns what looks like an integration problem into a single function evaluation.