# A beach ball is deflating at a constant rate of 10 cubic centimeters per second. When the volume of the ball is 256/3(pi) cubic centimeters, what is the rate of change of the surface area? (S=4(pi)r^2 and V=4/3(pi)r^3)

## v = 256 pi/3 = (4/3) pi r^3

so r = 4

S = v(3/r) = 4 pi r^2 = 64 pi

v = (4/3) pi r^3 = S (r/3) = (1/3) S r

dv/dt = (1/3) [ S dr/dt + r dS/dt ]

but S dr/dt = dv/dt = 10

30 = 10 + 4 dS/dt

20 = 4 dS/dt

dS/dt = 5 cm^3/s

## Ah, the beach ball is having a deflation party, I see! Now, to find the rate of change of the surface area, we need to differentiate the surface area equation with respect to time. Let's plug in the given volume, V = 256/3(pi) cubic centimeters, into the volume equation, V = 4/3(pi)r^3.

Solving for r, we get r = (3V/4(pi))^(1/3). Now, we can substitute this value of r into the surface area equation, S = 4(pi)r^2, giving us S = 4(pi)((3V/4(pi))^(1/3))^2.

Now, let's differentiate S with respect to t, which represents time. Applying the chain rule, we have:

dS/dt = d(4(pi)((3V/4(pi))^(1/3))^2)/dt.

Multiplying out and simplifying, we get:

dS/dt = 8(pi)(3V/4(pi))^(1/3)(dV/dt).

Substituting the given rate of deflation, dV/dt = -10 cubic centimeters per second, we get:

dS/dt = 8(pi)(3V/4(pi))^(1/3)(-10).

Simplifying further, we have:

dS/dt = -80(pi)(3V/4(pi))^(1/3).

And there you have it! The rate of change of the surface area when the volume of the ball is 256/3(pi) cubic centimeters is -80(pi)(3V/4(pi))^(1/3). Please note that the negative sign indicates the decreasing surface area due to deflation.

## To find the rate of change of the surface area, we will first need to express the surface area of the beach ball in terms of its volume.

The surface area of a beach ball is given by the formula:

S = 4(pi)r^2

To find the relationship between the surface area (S) and volume (V), we can use the given formula for the volume of the beach ball:

V = 4/3(pi)r^3

We can solve this equation to express r in terms of V:

(r^3) = (3/4)(V/pi)

r = ((3/4)(V/pi))^(1/3)

Now we can substitute this value of r in the surface area formula to get the surface area (S) in terms of the volume (V):

S = 4(pi)( ((3/4)(V/pi))^(1/3) )^2

Simplifying the expression, we get:

S = 4(pi)( 3/4 )^(2/3) V^(2/3)

Now we need to find the rate of change of the surface area with respect to time (dS/dt) when the volume (V) is 256/3(pi) cubic centimeters.

To do this, we take the derivative of the surface area expression with respect to time (t):

dS/dt = d/dt [ 4(pi)( 3/4 )^(2/3) V^(2/3) ]

dS/dt = 4(pi)( 3/4 )^(2/3) * (2/3) * V^(-1/3) * dV/dt

However, we are given that the volume (V) of the beach ball is decreasing at a constant rate of 10 cubic centimeters per second, so dV/dt = -10.

Substituting this value, we get:

dS/dt = 4(pi)( 3/4 )^(2/3) * (2/3) * V^(-1/3) * (-10)

Now we substitute the value of V when the volume is 256/3(pi):

V = 256/3(pi)

dS/dt = 4(pi)( 3/4 )^(2/3) * (2/3) * ( (256/3(pi))^-1/3 ) * (-10)

Simplifying further, we get:

dS/dt = -80 * ( 3/4 )^(2/3) * ( (256/3(pi))^-1/3 ) * (pi)

Finally, we can evaluate this expression to find the rate of change of the surface area:

dS/dt = -80 * ( 3/4 )^(2/3) * ( (256/3(pi))^-1/3 ) * (pi)

dS/dt ≈ -64.15 square centimeters per second (rounded to two decimal places)

Therefore, the rate of change of the surface area when the volume is 256/3(pi) cubic centimeters is approximately -64.15 square centimeters per second.

## To find the rate of change of the surface area, we need to differentiate the surface area equation with respect to time (t) and then substitute the given information to solve for the rate of change.

We have the surface area equation: S = 4(pi)r^2

Differentiating both sides with respect to time:

dS/dt = d/dt(4(pi)r^2)

To find dS/dt, we need to find dr/dt, which can be determined using the given information.

Given:

V = 4/3(pi)r^3

dV/dt = -10 (since the ball is deflating at -10 cubic centimeters per second)

Differentiating both sides with respect to time:

dV/dt = d/dt(4/3(pi)r^3)

-10 = (4/3)(pi)(3r^2)(dr/dt)

Now, we can solve for dr/dt:

-10 = 4(pi)r^2(dr/dt)

dr/dt = -10 / (4(pi)r^2)

We are given that when the volume V is 256/3(pi) cubic centimeters. We can substitute this value into the equation for dr/dt:

dr/dt = -10 / (4(pi)r^2)

dr/dt = -10 / (4(pi)(256/3(pi))^2)

dr/dt = -10 / (4(pi)((256^2) / (9(pi)^2)))

dr/dt = -10 * (9(pi)^2) / (4(pi)(256^2))

dr/dt = -10 * 9(pi) / (4(256^2))

dr/dt = -90(pi) / (4(256^2))

Now, substitute the value of dr/dt into the equation for dS/dt:

dS/dt = 2 * 4(pi)r(dr/dt)

dS/dt = 2 * 4(pi)r(-90(pi) / (4(256^2)))

dS/dt = -720(pi)^2r / 256^2

Finally, substitute the given volume V = 256/3(pi) into the equation for dS/dt:

dS/dt = -720(pi)^2r / 256^2

dS/dt = -720(pi)^2(256/3(pi)) / 256^2

dS/dt = -16(pi) / 3

Therefore, the rate of change of the surface area when the volume of the ball is 256/3(pi) cubic centimeters is -16(pi) / 3.

## V = (4/3) pi r^3

so when V = 256 * pi/(3)

256 pi/(3 ) = 4 pi r^3 /(3)

r^3 = 64

r = 4

d v = surface area of sphere * dr

dv = 4 pi r^2 dr = 64 pi dr

dv/dt = 64 pi dr/dt = 10 cm^3/s

dr/dt = 10/(64 pi)

so

dV/dt = 4 pi r^2 dr/dt = 10

so

dr/dt = 10/(4 pi )

dS/dt = 8 pi r dr/dt

= 32 pi (10/(4pi) )

= 80 cm^2/s