The sun is shining and a spherical snowball of volume 210 ft3 is melting at a rate of 14 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 3 hours?

I posted this question before and when i went back and did the math i got

0.09508332786
but it says its wrong

opps never mind i got it ... its supposed to be negative that number

To find the rate at which the radius is changing, we can use the volume formula for a sphere:

V = (4/3)πr^3,

where V is the volume and r is the radius.

Given that the initial volume V = 210 ft^3 and the rate of change of volume dV/dt = -14 ft^3/hr (negative because the snowball is melting), we can differentiate both sides of the volume formula with respect to time:

dV/dt = (4/3)π(3r^2)(dr/dt).

Now we can plug in the given values and solve for dr/dt:

-14 = (4/3)π(3r^2)(dr/dt).

First, let's solve for r when t = 3:

210 = (4/3)πr^3,
r^3 = (3/4)(210/π),
r ≈ 5.08 ft.

Now, we can solve for dr/dt when r ≈ 5.08 ft and dV/dt = -14 ft^3/hr:

-14 = (4/3)π(3(5.08)^2)(dr/dt).

Simplifying:

-14 = (4/3)π(3(5.08)^2)(dr/dt),
-14 = (4/3)π(3 * 25.8064)(dr/dt),
-14 ≈ (4/3)π(77.4192)(dr/dt),
-14 ≈ (4/3)(77.4192)(dr/dt),
-14 ≈ 154.8384(dr/dt),
dr/dt ≈ -14/154.8384,
dr/dt ≈ -0.090228.

Therefore, the rate at which the radius is changing after 3 hours is approximately -0.090228 ft/hr.

It seems like your initial calculation of approximately 0.09508332786 ft/hr was correct, but it may have been marked wrong due to rounding or incorrect submission format.

To find the rate at which the radius is changing, we need to use the formula for the volume of a sphere and differentiate it with respect to time.

The volume of a sphere can be expressed as:

V = (4/3)πr^3

Where V is the volume and r is the radius.

To find the rate at which the radius is changing, we need to differentiate both sides of the equation with respect to time (t):

dV/dt = d/dt[(4/3)πr^3]

Since the volume is changing with time, we can express dV/dt as the rate at which the volume is changing, which is given as -14 ft^3/hr:

-14 = d[(4/3)πr^3]/dt

To find the rate at which the radius is changing, we need to solve for dr/dt:

dr/dt = (-14) / [(4/3)πr^2]

Now, we know that after 3 hours, the volume of the snowball is 210 ft^3. We can use this information to find the radius (r) after 3 hours.

V = (4/3)πr^3
210 = (4/3)πr^3

Now we can solve for r:

r = ((3V) / (4π))^(1/3)
r = ((3 * 210) / (4π))^(1/3)

Now plug in the given values and calculate r:

r ≈ 3.894 ft

Now, substitute this value of r and the given rate of change (-14 ft^3/hr) into the equation for dr/dt to find the rate at which the radius is changing after 3 hours:

dr/dt = (-14) / [(4/3)π(3.894)^2]

Simplifying this expression, you should get:

dr/dt ≈ -0.036 ft/hr

Therefore, the rate at which the radius is changing after 3 hours (rounded to 5 decimal places) is approximately -0.036 ft/hr.