Help check a related rate problem

[neshepard]

Homework Statement

Before anybody tells me to check the web for this problem, I have and this is very different from what is out there.

An oil spill spreads in a circular pattern. Suppose the area is increasing at a rate of 800 square feet for every 1 foot increase in its radius (which is increasing at a rate of 15 feet per hour.) What is the rate of change of the area of the spill with respect to time?

Homework Equations

The Attempt at a Solution

So I am looking for somebody to explain why my answer is wrong and the professors is right (and nobody needs to say because he's the professor). I asked and he couldn't explain this clearly and I am very curious. This was on our last test.

My attempt:
A=πr^2
A'=2πr
dr/dt=15 da/dr=800 find da/dt
da/dt=2*π*800*15
=24000π ft^2/hr

His answer:
da/dt=800*15
=12000 ft^2/hr

Why is his right and where is the 2π?

 

[Mark44]

Homework Statement

Before anybody tells me to check the web for this problem, I have and this is very different from what is out there.

An oil spill spreads in a circular pattern. Suppose the area is increasing at a rate of 800 square feet for every 1 foot increase in its radius (which is increasing at a rate of 15 feet per hour.) What is the rate of change of the area of the spill with respect to time?

Homework Equations

The Attempt at a Solution

So I am looking for somebody to explain why my answer is wrong and the professors is right (and nobody needs to say because he's the professor). I asked and he couldn't explain this clearly and I am very curious. This was on our last test.

My attempt:
A=πr^2
A'=2πr
dr/dt=15 da/dr=800 find da/dt
da/dt=2*π*800*15
=24000π ft^2/hr

His answer:
da/dt=800*15
=12000 ft^2/hr

Why is his right and where is the 2π?

For this problem it's best not to use primes to indicate derivatives, because the prime doesn't indicate which variable your differentiation with respect to. IOW, in this equation you wrote, A'=2πr, does A' mean dA/dt or dA/dr?

From your work,
dA/dr = 800 ft²/ft
dr/dt = 15 ft/hr

So dA/dt = dA/dr *dr/dt = 800 ft²/ft * 15 ft/hr = 12,000 ft²/hr

 

[neshepard]

For this problem it's best not to use primes to indicate derivatives, because the prime doesn't indicate which variable your differentiation with respect to. IOW, in this equation you wrote, A'=2πr, does A' mean dA/dt or dA/dr?

From your work,
dA/dr = 800 ft²/ft
dr/dt = 15 ft/hr

So dA/dt = dA/dr *dr/dt = 800 ft²/ft * 15 ft/hr = 12,000 ft²/hr

I worked the problem with A' being da/dt since the question is how big is the area getting with respect to time. Since the problem gave my a growth of 800ft^2 for every 1 foot of radius, I have to make that da/dr (or area in relation to radius) and 15 ft per hour as dr/dt (since it's radius in relation to time). But according to the professor, we can just throw out the constants 2π? Since I have nothing that can account for them, how is that possible?

Thanks for the help.

 

[vela]

So I am looking for somebody to explain why my answer is wrong and the professors is right (and nobody needs to say because he's the professor). I asked and he couldn't explain this clearly and I am very curious. This was on our last test.

My attempt:
A=πr^2
A'=2πr
dr/dt=15 da/dr=800 find da/dt
da/dt=2*π*800*15

What was your thinking when you wrote down the last line? Just curious.

EDIT: I see you posted while I was editing, but even if you thought A'=da/dt, I don't see how you came up with that last line.

 

[Mark44]

I worked the problem with A' being da/dt
If A = πr², then A' = 2πr means dA/dr. You have differentiated with respect to r to get 2πr.

The assumption in this problem is that A is a function of r, and r is a function of t, so A is also a function of t.

The relationship is dA/dt = dA/dr * dr/dt, something I stated in my earlier post.

There is no point in calculating dA/dr, since you are given what it is; namely, 800 ft²/ft. The 800 ft² is already the area of a circular region, so you don't need to find its radius and calculate it.

since the question is how big is the area getting with respect to time. Since the problem gave my a growth of 800ft^2 for every 1 foot of radius,

Again, this is your dA/dr.

I have to make that da/dr (or area in relation to radius) and 15 ft per hour as dr/dt (since it's radius in relation to time). But according to the professor, we can just throw out the constants 2π? Since I have nothing that can account for them, how is that possible?

Thanks for the help.

The constant 2π didn't get thrown out because it didn't enter into things in the first place. If I asked you what is the area of a circle whose area is 10 ft², would you try to find the radius and then compute the area from that?

One other thing. You started with A for area. Don't switch to a to mean the same thing as A.

 

[neshepard]

What was your thinking when you wrote down the last line? Just curious.

EDIT: I see you posted while I was editing, but even if you thought A'=da/dt, I don't see how you came up with that last line.

vela-Probably why my problem is wrong is what the last line means..lol!

(As an aside, I didn't switch the a's for A's or anything, i don't type as fast as i think and I will sometimes not Capitalize A letter is all that was...but it does seem that given what I wrote, i would mean the same thing with dA/dt and da/dt (or DR/dt and dr/dt) etc.)

I appreciate the help from Mark44 though and the above is not meant as an attack or anything, just to let you know.

 

[Mark44]

My point about the capitalization is that if you get sloppy and aren't consistent with your use of variables, it will come back around and bite you when a problem has, say, both a and A in it. Or r and R, as did a problem posted here in the last week.