4. At time t, t>0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = 0, the radius of the sphere is 1 and at t = 15 the radius is 2.

a. Find the radius of the sphere as a function of t.
b. At what time t will the volume of the sphere be 17 times its volume at t = 0?

its your moma

recall that the voulme of a sphare of radius r is v(r)=(4pir^3)/3. find l, ther linearisation of v(r) at r=50. A sphare of radius 50 centimeters is covered with a layer of point of thickness 0.31 millimeters. use the linearisation of v at r=50 to estimate the volume of point that's used.

a. To find the radius of the sphere as a function of time, we can start by setting up a differential equation based on the given information.

Let V(t) be the volume of the sphere at time t, and r(t) be the radius of the sphere at time t.

We are given that the volume of the sphere is increasing at a rate proportional to the reciprocal of its radius. Mathematically, this can be expressed as:

dV/dt = k * (1 / r)

where k is the proportionality constant.

We also know that the volume of a sphere is given by the formula:

V = (4/3) * π * r^3

Differentiating the volume formula with respect to time, we have:

dV/dt = (4/3) * π * 3r^2 * dr/dt

Since dV/dt = k * (1 / r), we can equate the two expressions and solve for dr/dt:

(4/3) * π * 3r^2 * dr/dt = k * (1 / r)

Canceling out common terms and rearranging, we get:

dr/dt = (k / (4π * r^2))

Now, we can solve this separable differential equation to find the relationship between r and t.

Separating the variables, we have:

(4π * r^2) * dr = k * dt

Integrating both sides, we get:

∫(4π * r^2) * dr = ∫k * dt

(4π * (r^3) / 3) = k * t + C

Simplifying, we have:

r^3 = (3k * t + 3C) / (4π)

At t = 0, the radius of the sphere is 1. Substituting these values into the equation, we have:

1 = (3k * 0 + 3C) / (4π)

1 = 3C / (4π)

C = (4π) / 3

Substituting the value of C back into the equation, we have:

r^3 = (3k * t + 3(4π/3)) / (4π)

Simplifying further, we have:

r^3 = (3k * t + 4π) / (4π)

Taking the cube root of both sides, we get:

r = ((3k * t + 4π) / (4π))^(1/3)

Thus, the radius of the sphere as a function of t is given by:

r(t) = ((3k * t + 4π) / (4π))^(1/3)

b. Now we need to find the time t at which the volume is 17 times its volume at t = 0.

The volume of the sphere at t = 0 is:

V(0) = (4/3) * π * (1^3)
= (4/3) * π

We want to find t such that the volume V(t) is 17 times V(0). Mathematically, this can be expressed as:

V(t) = 17 * V(0)

Substituting the expressions for V(t) and V(0), we have:

(4/3) * π * r^3 = 17 * (4/3) * π

Canceling out common terms, we get:

r^3 = 17

Using the expression for the radius of the sphere as a function of t derived earlier, we can solve for t when r^3 = 17:

((3k * t + 4π) / (4π))^(1/3) = 17^(1/3)

Raise both sides to the power of 3 to eliminate the cube root:

3k * t + 4π = 17

Simplifying, we find:

3k * t = 17 - 4π

t = (17 - 4π) / (3k)

So, at time t = (17 - 4π) / (3k), the volume of the sphere will be 17 times its volume at t = 0.

To answer these questions, we can use the concept of related rates. We are given that the volume of the sphere is increasing at a rate proportional to the reciprocal of its radius.

Let's start by finding the radius of the sphere as a function of t (question a).
Let V be the volume of the sphere and r be the radius at time t. We know that the volume of a sphere is given by the formula V = (4/3)πr^3.

We are given that the volume is increasing at a rate proportional to the reciprocal of the radius, so we can write this relationship as:
dV/dt = k * (1/r)
where k is the constant of proportionality.

To find the radius as a function of t, we need to separate variables and integrate both sides of the equation.

dV/dt = k * (1/r)
dV = k * (1/r) * dt
4πr^2 * dr = k * dt (since dV = 4πr^2 * dr and dt = dt)

Now, we can integrate both sides:

∫ 4πr^2 * dr = ∫ k * dt
(4/3)πr^3 = kt + C
r^3 = (3kt + 3C) / (4π)
r = [(3kt + 3C) / (4π)]^(1/3)

Next, we can solve for the unknown constant C by using the given initial condition at t = 0, where the radius is 1.

r(0) = [(3k(0) + 3C) / (4π)]^(1/3)
1 = [(3C) / (4π)]^(1/3)

Cancelling out the cube root, we have:
1 = (3C) / (4π)

Solving for C:
C = (4π) / 3

Substituting this value of C back into our expression for r, we get:

r = [(3kt + 4π) / (4π)]^(1/3)

Now, let's move on to question b.

We need to find the value of t when the volume of the sphere is 17 times its volume at t = 0.

Using the volume formula V = (4/3)πr^3, we can set up the equation:

V(t) = 17 * V(0)

Substituting the expression we found for r into the volume formula, we get:

(4/3)π[(3kt + 4π) / (4π)]^(1/3)^3 = 17 * (4/3)π(1^3)

Simplifying, we have:

(3kt + 4π)^(1/3) = 17^(1/3)

Raise both sides to the power of 3 to eliminate the cube root:

(3kt + 4π) = 17

Now, solve for t:

3kt = 17 - 4π
t = (17 - 4π) / (3k)

Therefore, the volume of the sphere will be 17 times its volume at t = 0 when t = (17 - 4π) / (3k).