It's one-thirty am and Calculus is frying my brain... Help?

So you probably know about using similar triangles for this sort of problem.

We have one triangle formed by the pole, the distance from the base of the pole

to the tip of the man's shadow and the resulting hypotenuse.

The second smaller but similar triangle is formed by the man, the distance from

the man's feet to the tip of his shadow and the resulting hypotenuse.

Letting the distance from the base of the pole to the man's feet be x and the

distance from the man's feet to the tip of his shadow be s, then the similarity

of the two triangles tells us that

(x + s)/20 = s/6 ----> 6*(x + s) = 20*s ---> 6x = 14s ----> s = (3/7)*x.

So now taking the derivative of both sides of this last equation with respect to time,

and noting that the man is walking away at a speed of dx/dt = 5 ft/s, we get

ds/dt = (3/7)*dx/dt = (3/7)*5 = (15/7) ft/s, as you have found.

For B.), what you need to calculate is the rate of change of the tip of his shadow

relative to the base of the pole, i.e.,

d/dt(x + s) = dx/dt + ds/dt = 5 + (15/7) = 50/7 ft/s.

Part (b) is somewhat of a "trick" question, but it's a good one!

Let D be the distance between the streetlight and the man at time t, and let s be the distance between the man and the end tip of his shadow at time t.

Then, the distance between the streetlight and the tip of the man's shadow, at time t, is given simply by the sum of the previous two distances, or D + s.

(D + s) ' (t) is the quantity we want. Since (D + s) ' (t) = D '(t) + s '(t), and we know that D '(t) = 5ft/s and s '(t) = 15/7 ft/s, we know (D + s) '(t) = 5+15/7 = 50/7 ft/s or 7.14ft/s.

The key to part (b) is realizing that the speed of the tip of the man's shadow is contributed to by two speeds, one of which is constant in time and one of which varies in time. Specifically, the man's walking speed is constant, at 5ft/s, but the rate of the lengthening of the man's shadow is time-dependent. The sum of these two quantities determines how fast the tip of the shadow is moving at any time t.

If you drew a diagram for this problem, the quantity to be solved for in part (b) would be the rate at which the base length of the largest right triangle increases.

Assuming you're correct about the shadow lengthening at 2.14 feet per second, the tip of the shadow is moving at 7.14 feet per second at that point, because the base of the shadow is moving at the same speed as the man, which is 5 feet per second.

pole-height = 20 ft,

man-distance = 40 ft,

Let,

shadow length = x ft.

For similar triangles : (40+x)/x = 20/6,

OR,

6(40+x) = 20x, ===> 240 +6x = 20x,

OR,

20x -6x = 240, ===> 14x = 240,

OR,

x = 240/14 = 120/7 ft. ................................................. [1]

Man's speed = 5 ft/sec

and

Let,

shadow speed = s ft/sec,

SO,

(40+5+s)/s = 20/6,

6(45+s) = 20s, ===> 270 +6s = 20s,

20s-6s = 270,

14s = 270, ===> s = 270/14,

Thus,

shadow-speed = s = 19.29 ft/sec >=====================< ANSWER

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<----- x ----><-- y -->

If he is x feet away from the streetlight, and his shadow is y feet long,

then 6/y = 20/(x + y)

6x + 6y = 20y

6x = 14y

y = (3/7)x

(A)

dy/dt = (3/7)dx/dt = 15/7 ft/s

(B)

The tip of his shadow is at x + y.

So it is moving at dx/dt + dy/dt = 5 ft/sec + 15/7 ft/sec = 7 1/7 ft/sec