Help needed with Derivative Problems.?

Question

I need help in figuring out the answers to these two questions.

1. A vertical circular cuylinder has radius r feet and height h feet. If the height and radius both increase at the constant rate of 2ft/sec, then the rate, in square feet per second, at which the lateral surface area increases is: Answer Choices: A.4(pie)r B.2pie(r+h) C. 4pie(r + h) D. 4(pie)(r)(h) E. 4(Pie)(h)


2.Two cars are traveling along perpendicular roads, car A at 40 mi/hr and car B at 60 mi/hr. At noon when car A reaches the intersection, car B is 90 mi away, and moving toward at it. At 1 P.M. the distance between cars is changing, in miles per hour, at the rate of: Answer Choices: A. -40 B. 68 C.4 D.-4 E. 40


Thank you so much in reading this and helping me out! Its greatly appreciated!!


:Erin

Answer 1

1. The lateral surface area is
L = 2πrh
dL/dt = 2π (h dr/dt + r dh/dt)
dL/dt = 2π (2h + 2r) = 4π (h+r)


2. x² + y² = z²
x = car A , y = car B, z = distance between them

2x dx/dt + 2y dy/dt = 2z dz/dt

now x = 40 , y = 30 thus z = 50
dx/dt = 40 , dy/dt = -60

... you can plug this in...
Hint: the answer is negative...


§

Answer 2

First of all, it's not "pie", it's pi. :) This said, here are your problems:

1. None of the answers makes any sense. Since both r and h increase with time, the rate at which the lateral area increases must be a function of time, not of r or h, which are both functions of time.

The circumference of a circle is 2*pi*r, so the lateral surface of a cylinder is S = 2*pi*r*h. Given that r = 2t and h = 2t,

S = 8 * pi * t^2

so S' = 16 * pi * t

where t is time in seconds. Mathematically, it is equivalent to 4*pi*(r+h), but it still makes no sense to define one dependent variable through two others...

2. Since the roads are perpendicular, you can use the Pythagorean theorem to compute the distance between two cars:

d^2 = (40*t)^2 + (90 - 60*t)^2

Find d as a function of t, differentiate it, and you should have your answer as the derivative at t = 1.

On a second thought, though, DC's solution above is much easier...