1. a Spherical snowball melts in such a way that the instant at which its radius

Question

1. a Spherical snowball melts in such a way that the instant at which its radius is 20 cm, its radius is decreasing at 3 cm/min. At what rate is the volume of the ball of snow changing at that instant?

2. A ladder 25 feet long is leaning against the wall of the house, while the base of the ladder is pulled away from the wall at a rate of 3 feet per second.a. How fast is the top moving down the wall when the base of the ladder is 20 feet away from the wall? (b) How fast is the angle between the ladder and the wall changing when the base of the ladder is 15 feet away from the wall?

Explanation / Answer

1.Let radius be r,
Volume V = (4/3)(pi)r^3
Differentiating,
dV/dr = (4/3)(pi)(3r^2) = 4(pi)(r^2)
Also dV/dr = (dV/dt) / (dr/dt),
So dV/dt = 4(pi)(r^2)(dr/dt),
At r=20 cm, dr/dt=-3cm/min, putting these in above equation,
dV/dt = 4(pi)[(20cm)^2](-3cm/min)
dV/dt =- 4800(pi) (cm^3)/min

2. l² = x² + y² : rightside is positive for x and upside is positive for y
d(l²) = d(x²) + d(y²)
0 = 2xdx + 2ydy
dy = -(x/y)dx
dy/dt = -(x/y)dx/dt : Eq1
we have dx/dt = +3 ft/sec

a) her l =25 length of ladder.

x= 20 ft then y = (l² - x²) = (25² - 20²) = +15 ft

dy/dt = -(x/y)dx/dt = -(20/15)(2) = -8/3 -2.667 ft/sec

b)
x = +15 ft then y = +20 ft
cos = y/l = y/25
sin = x/l = x/25
dsin = d(x/25)
cos d = dx/25
y/25 d = dx/25
d = dx/y
d/dt = (dx/y)/dt
d/dt = (dx/dt)/y
d/dt = (+2)/20 = +0.1 rad/sec

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