Calc related rates problems (2)?

Question

1) A street light is mounted at the top of a 15 ft pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole.
2) Water is leaking out of an inverted conical tank at a rate of 10,000 cm cubed/minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into a tank.

Answer 1

Only one problem per question please.

2)
Let
V = volume water
h = height water
r = radius at surface of water

Given
dh/dt = 20 cm/min
water is leaking out at 10,000 cc/min

Find dV/dt when h = 2m = 200 cm.

First solve for dV/dt without the leak and then add 10,000.

From the shape of the tank we have:

h/r = 6/(4/2) = 6/2 = 3
h = 3r
r = h/3

V = (1/3)πr²h = (1/3)π(h/3)²h = πh³/27

Take the derivative.

dV/dh = πh²/9

dV/dt = (dV/dh)(dh/dt) = (πh²/9)(20) = 20πh²/9
dV/dt = 20πh²/9 = 20π(200)²/9 = 800,000π/9 cc/min

But water is leaking out at a rate of 10,000 cc/min, so taking that into account:

dV/dt = 800,000π/9 + 10,000 cc/min ≈ 289,253 cc/min

Answer 2

you're searching for an equation that makes use of your variables and additionally you're instructed it rather is a precise triangle, so: x^2 + y^2 = 625 The length of the ladder is persevering with at 25' you're searching for dy/dt at x = 7, and given dx/dt = 2 So utilising the chain rule: 2xdx/dt + 2ydy/dt = 0 dy/dt = -(x/y)dx/dt while x = 7, y = 24 (Pythagorean Thm) so dy/dt = -(7/24)2 = -.583 ft/s b. Do the comparable way in common terms the equation is (one million/2)xy