Long difficult word problem for calculus H.W below.....?

Question

The upper end of a straight ladder is leaning against the vertical side of a building, with its lower end on a horizontal sidewalk. To the consternation of the painter standing on it, the upper end is slipping down the wall at a constant rate of 2 feet/sec. The lower end is, therefore, sliding along the sidewalk. Having had a course in the Calculus, the painter is, of course, interested in how fast the lower end is moving. If the ladder is 25 feet long, how fast is the lower end moving when the upper end is 7 feet above the ground?

a. 8/3 feet/sec.

b. 3/2 feet/sec

c. 48/7 feet/sec.

d. 7/12 feet/sec.

e. 15/4 feet/sec.

f. 16/15 feet/sec.

g. none of these

Answer 1

If you call the height of the ladder on the wall h, and the distance of the foot from the wall d, then they are related by the Pythagorean Identity:
h^2 + d^2 = 25^2

Using implicit differentiation:
d/dt(h^2 + d^2) = d/dt(625)
d/dt(h^2) + d/dt(d^2) = 0
2h * dh/dt + 2d * dd/dt = 0
h * dh/dt + d * dd/dt = 0

We are told that dh/dt = -2 ft/sec
We are interested in dd/dt when h = 7.
When h = 7, our original tells us d = sqrt(25^2 - 7^2) = sqrt(576) = 24.

So,
7 * -2 + 24 * dd/dt = 0
-14 + 24 * dd/dt = 0
24 * dd/dt = 14
dd/dt = 14/24 = 7/12 ft/sec

The answer is d.

Answer 2

The hypotenuse of your triangle is given: 25 ft
25^2 = v^2 + h^2
h is horizontal and v is vertical
Solve for v:
v = sqrt(625 - h^2)
Find dv/dh:
dv/dh = -h/sqrt(625 - h^2)
You're given the rate of change of v with respect to time:
dv/dt = -2
So (dv/dt)/(dv/dh) will give you dh/dt:
-2/(-h/sqrt(625 - h^2)) = 2sqrt(625 - h^2)/h = dh/dt
You know that currently v = 7
h = 625 - 7^2 = 24
Plug in h:
2sqrt(625 - 24^2)/24 = 7/12

D