Assume that the radius 𝑟 of a sphere is expanding at a rate of 12 in./min.

Question

Assume that the radius 𝑟 of a sphere is expanding at a rate of 12 in./min.

12 in./min.The volume of a sphere is 𝑉=4/3pi*r^3
Determine the rate at which the volume is changing with respect to time when 𝑡=7 min.assuming that 𝑟=4 @ 𝑡=0

Answer 1

r = 4 + 12t

v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt
at t=7,
dv/dt = 4π(4+12*7)^2 (12) = ___

Answer 2

Well, here we go!

First, let's plug in the values we have into the volume formula: V = (4/3)πr^3.

At t = 0, the radius (r) is 4 inches. So, when t = 0, r = 4.

Now, we need to find the rate at which the volume is changing with respect to time (dV/dt) at t = 7 minutes.

To do that, we can take the derivative of the volume function with respect to time (t). The derivative of V with respect to t is given by:

dV/dt = 4πr^2(dr/dt)

We already know that dr/dt is 12 inches/min.

So, let's plug in the values we know:

r = 4 (at t = 7 min)
dr/dt = 12 inches/min

dV/dt = 4π(4^2)(12)
= 4π(16)(12)
= 768π

Therefore, when t = 7 minutes, the rate at which the volume is changing is 768π cubic inches per minute.

But hey, what's π doing in the gym? It's getting circumference-jacked!

Hope that answer didn't fall flat, and remember, math can be fun!

Answer 3

To find the rate at which the volume is changing with respect to time, we can use the chain rule of differentiation.

Given:
- The radius of the sphere is expanding at a rate of 12 in./min.
- The volume of a sphere is given by 𝑉 = (4/3)π𝑟³.
- At 𝑡 = 0, the radius 𝑟 = 4.

Let's find the rate of change of volume 𝑉 with respect to time 𝑡 when 𝑡 = 7 min.

Step 1: Find the derivative of volume 𝑉 with respect to the radius 𝑟.
Taking the derivative of 𝑉 with respect to 𝑟:
𝑑𝑉/𝑑𝑟 = 4π𝑟²

Step 2: Find the rate of change of radius 𝑟 with respect to time 𝑡.
Given that 𝑟 is expanding at a rate of 12 in./min, we have:
𝑑𝑟/𝑑𝑡 = 12 in./min

Step 3: Apply the chain rule.
Since 𝑑𝑉/𝑑𝑡 = (𝑑𝑉/𝑑𝑟) * (𝑑𝑟/𝑑𝑡), we can substitute the derivatives found in Steps 1 and 2:
𝑑𝑉/𝑑𝑡 = (4π𝑟²) * (12 in./min)

Step 4: Substitute the values.
When 𝑡 = 7 min, 𝑟 = 4 (given at 𝑡 = 0). Let's substitute these values into the equation found in Step 3:
𝑑𝑉/𝑑𝑡 = (4π(4)²) * (12 in./min)

Step 5: Calculate the final result.
Compute the expression to get the rate of change of volume with respect to time:
𝑑𝑉/𝑑𝑡 = (4π(16)) * (12 in./min)
𝑑𝑉/𝑑𝑡 = 768π in³/min

Therefore, the rate at which the volume is changing with respect to time when 𝑡 = 7 min is 768π cubic inches per minute.

Answer 4

To determine the rate at which the volume is changing with respect to time, we can use the chain rule from calculus.

1. First, let's find an expression for the volume of the sphere as a function of time. We are given that the radius of the sphere is expanding at a rate of 12 in./min, so we can express the radius as a function of time, r(t). Since r(0) = 4, we can write r(t) = 4 + 12t, where t represents time in minutes.

2. Now, substitute the expression for r(t) into the volume formula V = (4/3)πr^3 to get the volume as a function of time: V(t) = (4/3)π(4 + 12t)^3.

3. To find the rate at which the volume is changing with respect to time, we need to differentiate V(t) with respect to time (t). Applying the power rule and chain rule, we have: dV/dt = (4/3)π * 3 * (4 + 12t)^2 * 12.

4. Simplify the expression to obtain the rate at which the volume is changing with respect to time: dV/dt = 16π(4 + 12t)^2.

5. To find the rate at which the volume is changing when t = 7 minutes, substitute t = 7 into the expression: dV/dt = 16π(4 + 12(7))^2.

6. Calculate the numerical value: dV/dt = 16π(4 + 84)^2 = 16π(88)^2.

7. Finally, evaluate the expression: dV/dt ≈ 493,007.47 in.^3/min.

Therefore, when t = 7 minutes, the rate at which the volume is changing with respect to time is approximately 493,007.47 in.^3/min.