(i) Related rates problmes: A) The length of a rectangle is increasing at a rate

Question

(i) Related rates problmes:

A) The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is 15 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

B) If a snowball melts so that its surface area decreases at a rate of 4 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm.

C) A plane flying horizontally at an altitude of 2 mi and a speed of 510 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station. (Round your answer to the nearest whole number.)

Explanation / Answer

solution-1 Area is a function of the length and width. Since length and width are changing with time, the area changes with time as well.
A(t) = l(t)*w(t)
Dropping the (t) notation,
A = lw
Implicity differentiating with respect to time,
A' = lw' + wl'
From the problem statement:
l = 15 cm
l' = 4 cm/s
w = 5 cm
w' = 6 cm/s
A' = (15 cm)(6 cm/s) + (5 cm)(4 cm/s)
= 90 cm²/s + 20 cm²/s = 110 cm²/s

solution - 2   Let S be the surface area, r the radius and x the diameter.
Then S = 4.r² = .d²
dS/dt = d/dt (.d²) = .2d.dx/dt (chain rule)
But we're told that dS/dt = -4, and that x = 9, so:
-4 = .(2*9)*dx/dt
so dx/dt = -4/(18) = -2/(9)
diameter decreases by 0.07071 cm/s

solution-3 let r :shortest distance between plane and station.
x:horizontal distance.
dr/dt=?
Let x = 83. Then
r = (2² + 3²) = (13)

r^2 = 2^2 + x^2
d/dt >> d(r^2)/dt = d(x^2)/dt => 2r dr/dt = 2x dx/dt
So
(dr/dt)|x=3 = x/r * dx/dt
= 3*510/sqrt(13) = 424.34 mi/h

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