# Thread: Tangents and the Derivative of V=(4/3)* π * r^3 at a point (r = 4)

1. ## Tangents and the Derivative of V=(4/3)* π * r^3 at a point (r = 4)

What is the rate of change of the volume of a ball
V=(4/3)* π * r^3

with respect to the radius when the radius is equals=4​?

The volume changes at a rate of?
(Type an exact answer, using pi as needed.)

So i started with
r = 4, insert into the equation.
V=(4/3)*pi*r^3
V= (4/3)*pi*(4)^3
V= 4/3*pi* 64

Then, slope of the curve at point (r,V)
lim h->0 f(r+h)-f(r)/h

=lim h->0 (f(4+h) - (4/3*pi*64))/h

=lim h->0 ((4/3*pi (4+h)^3) -
(4/3*pi*64))/h

am I on the right track?

2. What you've done so far is fine, and if you continue to simplify and evaluate the limit you will get the correct answer. But I think it would be way easier to differentiate first, and then plug in the point r = 4. What do you get when you calculate dV/dr?

3. Originally Posted by ksdhart2
What you've done so far is fine, and if you continue to simplify and evaluate the limit you will get the correct answer. But I think it would be way easier to differentiate first, and then plug in the point r = 4. What do you get when you calculate dV/dr?
Well if continue as started. My lim is this big long expression. And I'm lost on how to simplify it.
So I will try to differentiate first as you suggested.

Thanks

4. Originally Posted by muzik
lim h->0 ((4/3*pi (4+h)^3) - (4/3*pi*64))/h
Do you know the property of limits that allows you to factor out a constant and move it to the front of the limit statement?

EG: limit[3x - 3] is the same as 3*limit[x - 1]

Your result above contains a constant value that can be factored out. If you apply the property I mentioned, and then expand what's left in the numerator, you can then simplify further (dividing by h) and evaluate the entire expression (constant times limit).

Personally, I would have taken kdshart2's approach (determine the general derivative first), but each approach is valid.

5. Originally Posted by muzik
My [limit statement] is this big long $\text{e}$$\text{quation}$.
Here's a friendly note about math terminology: your limit statement (as posted) is an expression, not an equation. All equations contain an equals sign.

6. Originally Posted by mmm4444bot
Do you know the property of limits that allows you to factor out a constant and move it to the front of the limit statement?

EG: limit[3x - 3] is the same as 3*limit[x - 1]

Your result above contains a constant value that can be factored out. If you apply the property I mentioned, and then expand what's left in the numerator, you can then simplify further (dividing by h) and evaluate the entire expression (constant times limit).

Personally, I would have taken kdshart2's approach (determine the general derivative first), but each approach is valid.
I am going to try and determine the derivative first. then input the r value after. I didn't realize i could of factored out the constant. I'll see if that helps me on my next approach. I'm working it out now.

Thank you for the suggestion

7. Originally Posted by mmm4444bot
Here's a friendly note about math terminology: your limit statement (as posted) is an expression, not an equation. All equations contain an equals sign.
You're right. thanks

8. Ahhh so much better to determine the derivative first.

The volume changes at a rate of 4/3*pi*48. Thanks all for you help.

9. Originally Posted by muzik
The volume changes at a [rate] of 4/3*pi*48 [when the radius is 4 units].
Good job.

If you have time and motivation, I'd suggest that you simplify your original limit statement (factor out the constant, and apply the limit property), to confirm that you get the same end result. It's good practice.

10. Originally Posted by mmm4444bot
Good job.

If you have time and motivation, I'd suggest that you simplify your original limit statement (factor out the constant, and apply the limit property), to confirm that you get the same end result. It's good practice.
Thank you!!
And thanks for the challenge. When I do get some time I'll give it a shot. Now that I know the answer. I need all the practice I can get. I got a test on Thursday but got to finish up on homework first.

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