You are blowing air into a spherical balloon at a rate of 7 cubic inches per second. The goal of this problem is to answer the following question: What is the rate of change of the surface area of the balloon at time t= 1 second, given that the balloon has a radius of 3 inches at that instant?

Question

You are blowing air into a spherical balloon at a rate of 7 cubic inches per second. The goal of this problem is to answer the following question: What is the rate of change of the surface area of the balloon at time t= 1 second, given that the balloon has a radius of 3 inches at that instant?

(a) Next write a formula relating the changing volume V(t) of the sphere to the changing radius r(t), and differentiate that formula with respect to t. Using what you know about V'(t) and r(1), find the rate of change of the radius at t=1 sec:

(b) Finally, write a formula relating the changing surface area S(t) of the sphere to the changing radius r(t), and differentiate that formula with respect to t. Use what you know about r(1) and r'(1) to determine the rate of change of the surface area at t=1 sec:

Answer 1

a = 4πr^2

da/dt = 8πr dr/dt

Answer 2

(a) To relate the changing volume V(t) of the sphere to the changing radius r(t), we use the formula for the volume of a sphere, which is V = (4/3)πr^3.

Differentiating this formula with respect to t, we get dV/dt = 4πr^2(dr/dt).

Given that the balloon has a radius of 3 inches at t=1 second (r(1) = 3), we can substitute this value into the formula:

dV/dt = 4π(3^2)(dr/dt)
dV/dt = 36π(dr/dt)

(b) To relate the changing surface area S(t) of the sphere to the changing radius r(t), we use the formula for the surface area of a sphere, which is S = 4πr^2.

Differentiating this formula with respect to t, we get dS/dt = 8πr(dr/dt).

Using what we know about r(1) and r'(1):

dS/dt = 8π(3)(dr/dt)
dS/dt = 24π(dr/dt)

So, the rate of change of the surface area at t=1 sec is 24π(dr/dt) cubic inches per second.

Answer 3

To answer these questions, we will use the formulas for the volume and surface area of a sphere:

(a) The volume formula of a sphere is given by V = (4/3)πr^3, where V(t) represents the changing volume of the sphere and r(t) represents the changing radius of the sphere.

Differentiating the formula V = (4/3)πr^3 with respect to t, we get:
V'(t) = 4πr^2 * r'(t)

Since r(1) represents the radius at t=1 second, we know that r(1) = 3 inches.

To find the rate of change of the radius at t=1 second, we need to evaluate r'(1).

(b) The surface area formula of a sphere is given by S = 4πr^2, where S(t) represents the changing surface area of the sphere and r(t) represents the changing radius of the sphere.

Differentiating the formula S = 4πr^2 with respect to t, we get:
S'(t) = 8πr * r'(t)

Since r(1) represents the radius at t=1 second and we know r(1) = 3 inches, we need to determine r'(1) to find the rate of change of the surface area at t=1 second.

Answer 4

To find the rate of change of the surface area at t=1 second, we will first need to find the rate of change of the radius with respect to time.

(a) Let's start by writing a formula relating the changing volume V(t) of the sphere to the changing radius r(t). The volume of a sphere is given by the formula:

V = (4/3)πr^3

To relate V(t) and r(t), we can differentiate both sides of the equation with respect to t using the chain rule:

dV/dt = d((4/3)πr^3)/dt

Next, we can differentiate the right side of the equation using the power rule:

dV/dt = 4π(d(r^3)/dt)

Now, let's differentiate r^3 with respect to t using the power rule:

dV/dt = 4π(3r^2)(dr/dt)

Simplifying this equation, we have:

dV/dt = 12πr^2(dr/dt)

We want to find the rate of change of the radius at t=1 sec, so we can substitute r(1) into the equation above. Given that the radius at t=1 sec is 3 inches, we have:

dV/dt = 12π(3^2)(dr/dt)

Substituting the values, we get:

dV/dt = 108π(dr/dt)

Now, we need to find the value of dr/dt. Unfortunately, we don't have enough information to determine this value from the given problem statement.

(b) To find the rate of change of the surface area at t=1 second, we need to write a formula relating the changing surface area S(t) of the sphere to the changing radius r(t). The surface area of a sphere is given by the formula:

S = 4πr^2

We can differentiate both sides of the equation with respect to t using the chain rule:

dS/dt = d(4πr^2)/dt

Next, let's differentiate the right side of the equation using the power rule:

dS/dt = 4π(2r)(dr/dt)

Simplifying this equation, we have:

dS/dt = 8πr(dr/dt)

We want to find the rate of change of the surface area at t=1 sec, so we can substitute r(1) and dr/dt into the equation above. Given that the radius at t=1 sec is 3 inches, and we don't have the value of dr/dt, we cannot determine the rate of change of the surface area at t=1 sec without additional information.