A have a question about related rates from my calculus class may someone help me?

Question

A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 4ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 12s?

Answer 1

"A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 4ft/s."

 

=> dr/dt = +4 ft/s

 

Taking the antiderivative: r(t) = 4 t + C, r(0) = 0 = C,

so r(t) = 4 t.

 

"How rapidly is the area enclosed by the ripple increasing at the end of 12s?"

 

=> What is dA/dt at 12 s where A(t) = pi (r(t))^2 ?

 

A(t) = pi (r(t))^2 is the familiar formula for the area of a circle, A = pi r^2, with A and r treated as functions of time.

 

dA/dt = pi (2 r(t) dr/dt) using the chain rule.

 

At 12 s, dA/dt = 2*pi*r(12)*4 = 2*pi*4*12*4 = 384 pi ft^2

Answer 2

So, Allison, the given data are: dr/dt = 4 (units of ft s^-1) . == equation 1

You also know an equation for the area of a circle: A = .pi. r^2 . == equation 2

So you can put these together: dA/dr = 2 .pi. r . (differentiation of equation 2 with respect to r) == equation 3

Multiply equation s 1 and 3 together: dA/dr dr/dt = 4 x 2 .pi.r = dA/dt = 8 .pi. r  == equation 4

Since after 12 s the ripple will have r = 4 ft/s x 12 s = 48 ft,

 then dA/dt at that time will have value 8 .pi. 48 = 384 .pi. ft^2/s (unless I've done my multiplication wrong).

 

By the way, you would also get the same result by considering the problem *geometrically* as the area covered by the ripple acting for a small amount of time at t=12 sec, when it has a circumference of 2 .pi. r = 96 .pi. ft. Then the small "slice" of circumference (annulus), unwrapped to make it straight, is length 96 .pi. ft, width 4 ft/s, or 384 .pi.  ft^2/s.

 

Somewhat more interesting, perhaps, would be your finding if you tried to measure the speed of travel of a single crest of the ripple. You'd find it's travelling at 8 ft/s, or twice as fast as the ripple as a whole is moving! If you get interested in why, look up "deep water wave propagation".

 

Answer 3

A'=?

dr=4

sec=s=12
r=4s
r=4(12)

r=48

A=πr²
A'=2πr•dr
A'=2π(48)(4)
A'=384π ft/sec