Related rates and snowball melting

[benEE2018]

Homework Statement

If a snowball melts so that its surface area decreases at a rate of 10 cm^2/min, find the rate at which the diameter decreases when the diameter is 11 cm.

Homework Equations

I don't know what i am doing wrong.

The Attempt at a Solution

A=4(pi)r^2
dA/dt= ((0*r^2)+(4(pi)*2r(dr/dt)) [used the product rule to differentiate with respect to time
dA/dt= 4(pi)*2r(dr/dt) [simplified version]
We know dA/dt= -10cm^2/min
Diameter=11 cm so the radius is 5.5
dA/dt=4(pi)2r(dr/dt)
dr/dt=(dA/dt)/4(pi)2r [isolate (dr/dt)]
dr/dt=(-10cm^2/min)/4(pi)*11
dr/dt=-10/44(pi) but the answer is incorrect any suggestions would be greatly appreciated thanks

 

[gopher_p]

Homework Statement

If a snowball melts so that its surface area decreases at a rate of 10 cm^2/min, find the rate at which the diameter decreases when the diameter is 11 cm.

Homework Equations

I don't know what i am doing wrong.

The Attempt at a Solution

A=4(pi)r^2
dA/dt= ((0*r^2)+(4(pi)*2r(dr/dt)) [used the product rule to differentiate with respect to time
dA/dt= 4(pi)*2r(dr/dt) [simplified version]
We know dA/dt= -10cm^2/min
Diameter=11 cm so the radius is 5.5
dA/dt=4(pi)2r(dr/dt)
dr/dt=(dA/dt)/4(pi)2r [isolate (dr/dt)]
dr/dt=(-10cm^2/min)/4(pi)*11
dr/dt=-10/44(pi) but the answer is incorrect any suggestions would be greatly appreciated thanks

You have found the rate at which the radius changes, $\frac{dr}{dt}$. If you look again, you'll see that is not what the question is asking you to find.

Edit: From what I can tell, you have done your version of the problem correctly; i.e. if the problem had asked you to find the rate at which the radius changes, you would have the correct answer.

 

[benEE2018]

is there another related formula that includes diameter instead of the radius that i am suppose to differentiaite instead of 4(pi)r^2?

 

[benEE2018]

the hint for the homework was a=4(pi)r^2 so i know i have to be using that equation to solve the problem

 

[gopher_p]

Well, you could use $D=2r$ in one of at least two ways. You can subtitute $\frac{D}{2}$ for $r$ in the hint equation to get $a=\pi D$ and use that for the related rates part. Or you can differentiate $D=2r$ with respect to $t$ to get $\frac{dD}{dt}=2\frac{dr}{dt}$ and use what you've discovered about $\frac{dr}{dt}$ to say something about $\frac{dD}{dt}$.

One of the key components of being able to successfully work these related rates problems is answering the question, "What is the rate that they're asking me to find, and how do I express that using math notation?" If the problem had said, "Find the rate at which the radius changes when diameter is $11$cm", I might first write that as $\frac{dr}{dt}|_{D=11}=?$, which would indicate that I would need some expression on the right whose only unknown is $D$. Or I could, as you did, realize that $r=5.5$ when $D=11$, which changes the problem to $\frac{dr}{dt}|_{r=5.5}=?$, proceed to find $\frac{dr}{dt}$ in terms of $r$ and other known quantities (like you did), and plug $5.5$ in for $r$ (like you did).

 

[benEE2018]

thank you very much i differentiated a=4(pi)r^2 with r=D/2 so that when i differentiated the equation i would end up with a dD/dt which was what the problem was asking for. thank you gopher_p for your help i really appreciated it and also your tip was very useful for helping me solve my other homework problems!

 

[SteamKing]

The Attempt at a Solution

A=4(pi)r^2
dA/dt= ((0*r^2)+(4(pi)*2r(dr/dt)) [used the product rule to differentiate with respect to time

The product rule is unnecessary in this case. A = constant * r^2, where constant = 4 pi.

It's the chain rule you want here.