The radius, r, of a sphere is increasing at a constant rate of 0.05 meters per second.
A. At the time when the radius of the sphere is 12 meters, what is the rate of increase in its volume?
B. At the time when the volume of the sphere is 36pi cubic meters, what is the rate of increase in its surface area?
C. Express the rate at which the volume of the sphere changes with respect to the surface area of the sphere (as a function of r).
Parts A and B I have figured out (A. = 28.8pi cubic meters per second; B. = 1.2pi squared meters per second) But I can't quite figure out C. How do I relate surface area and volume together into one equation???
Given:
dr/dt = 0.05 m/s
V = 4/3 πr³
dV/dt = 4πr² (dr/dt)
A = 4πr²
dA/dt = 8πr (dr/dt)
- - - - - - - - - - - - - - -
First question:
What is dV/dt when r = 12?
dV/dt = 4π(12 m)² (0.05 m/s)
dV/dt = 28.8π m³/s
dV/dt ≈ 90.478 m³/s
- - - - - - - - - - - - - - -
Second question:
When V = 36π, what is dA/dt?
Using volume formula,
36π = 4/3 πr³
27 = r³
r = 3
dA/dt = 8π(3m) (0.05 m/s)
dA/dt = 1.2π m²/s
dA/dt ≈ 3.77
- - - - - - - - - - - - - - -
Third question:
Find dV/dt in terms of A(r) (and I'm assuming only in terms of A(r)).
A(r) = 4πr²
dV/dt = 4πr² (dr/dt)
Since you're given dr/dt = 0.05, plug it in:
dV/dt = 4πr² (0.05)
Also, since A(r) = 4πr², just replace this term in dV/dt with A(r):
dV/dt = 0.05 A(r)
V = (4/3)pi r^3
dV/dr = 4 pi r^2 which happens to be the area
so
dV/dt = 4 pi r^2 dr/dt
the differential change of volume is the surface area * dr
PS
I wonder how you did parts a and b without knowing that
I did notice this when I took the derivative for part A of the volume. But it seemed too simple to just plug in this value, I wanted to make sure that I was certain about the placement. Thank you!!!
dr/dt = .05
v = (4/3)pi r^3
dv/dt = (4/3)pi(3) r ^2 (dr/dt) <---- this step is where the mistake was. you needed to have differentiated the volume with respect to time. to do this use the power rule on the right side. the derivative of (4/3)pi r^3 is (4/3)pi(3) r ^2 (dr/dt). use the power rule to get this and multiply by radius differentiated with respect to time. (quick power rule for r^3: bring the 3 to the front and diminish the 3 by 1, giving you 3r^2)
then plug in dr/dt, or r(t) as what you wrote and get (4/3) pi (3)(12)^2 (.05) as your answer
= 28.8 pi
---------------------------------------...
B.
A = 4 pi r^2
V = 4/3 pi r^3
dr/dt = .05
you need to find the change in surface area (i used "A" as the variable) at the moment when the volume is 36 pi. The only other given number you have is dr/dt= .05
so you have the right volume but you need to find what the radius was when the volume equaled 36 pi.
you plug in 36 pi into the equation for volume where "V" is (to solve for "r"):
36 pi = (4/3) pi r^3
divide pi from each side.. 36= (4/3)r^3
multiply each side by (3/4) to get rid of the (4/3).... 27 = r^3
find the cube root of 27 and get.... r = 3
now you know what the radius is equal to at the moment when the volume is 36 pi.
differentiate the equation for surface area now with respect to time: A = 4 pi r^2
dA/dt = 4 pi 2 r (dr/dt)
plug in your value for r = 3 and (dr/dt)= .05
dA/dt = 4 pi 2 r (dr/dt)
dA/dt = 4 pi 2 (3)(.05)
= 1.2 pi
The volume of sphere is increasing at the rate of 1200c.cm/sec. the rate increase in its surface area when the radius is 10cm is
Well, let's see if we can tackle part C by relating the volume and surface area of a sphere.
The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius. And the surface area is given by A = 4πr².
To relate the rate of change of volume to the surface area, we can take the derivative of the volume equation with respect to time (t) and divide it by the derivative of the surface area equation with respect to time.
So, let's get to the math!
We have V = (4/3)πr³, and differentiating both sides with respect to t, we get:
dV/dt = (4/3)(3πr²)(dr/dt)
Similarly, for A = 4πr², differentiating both sides with respect to t, we get:
dA/dt = 4π(2r)(dr/dt)
Now, to express the rate at which the volume of the sphere changes with respect to the surface area, we divide the derivative of volume by the derivative of surface area:
(dV/dt) / (dA/dt) = [(4/3)(3πr²)(dr/dt)] / [4π(2r)(dr/dt)]
Guess what? LUCKY US! The (dr/dt) term cancels out, and we end up with:
(dV/dt) / (dA/dt) = (4/3)r² / (2r)
Simplifying further, we have:
(dV/dt) / (dA/dt) = 2/3r
And there you have it! The rate at which the volume of the sphere changes with respect to the surface area is (2/3r).
I hope that helps, and that you find it as funny as I found working on it!
To relate the surface area and volume of a sphere together, we need to use the formulas for surface area and volume of a sphere. Let's start by finding these formulas:
The surface area of a sphere, A, is given by the formula:
A = 4πr^2
The volume of a sphere, V, is given by the formula:
V = (4/3)πr^3
Now that we have the formulas, we can differentiate both with respect to time (t) using implicit differentiation, as the radius is changing with time:
For the surface area, A, we differentiate both sides of the equation with respect to time to get:
dA/dt = d(4πr^2)/dt
dA/dt = 8πr(dr/dt)
For the volume, V, we differentiate both sides of the equation with respect to time to get:
dV/dt = d((4/3)πr^3)/dt
dV/dt = 4πr^2(dr/dt)
Now, let's express the rate at which the volume of the sphere changes (dV/dt) with respect to the surface area (dA/dt) by rearranging the equations:
dV/dt = 4πr^2(dr/dt)
dr/dt = (1/(4πr^2))dV/dt
Since we want to express the rate of change of volume with respect to the surface area, we divide the equation by 8πr:
(dr/dt)/(8πr) = (1/(32π^2r^3))dV/dt
Now, we can express the rate at which the volume of the sphere changes with respect to the surface area as a function of r:
dV/dA = (1/(32π^2r^3))
So, the rate at which the volume of the sphere changes with respect to the surface area is given by the function (1/(32π^2r^3)), where r is the radius of the sphere.