Use chain rule to find rate of area increase

1. Mar 28, 2013

EricPowell

1. The problem statement, all variables and given/known data
32) A stone dropped into a pond at time t=0 seconds causes a circular ripple that ravels out from the point of impact at 5 metres per second. At what rate (in square metres per second) is the area within the circle increasing when t=10?

2. Relevant equations
I need to use the chain rule dy/dx = dy/du x du/dx

3. The attempt at a solution
The area of a circle is A=∏r2
Differentiating this formula will tell me the rate at which the area increases for a specific radius
dA/dr=2∏r

The rate at which the radius increases is 5 metres per second, so I think that would be expressed as
dr/dt=5m/s

I need to find the rate of change of the area of the circle at 10 seconds. Using this chain rule, I think this would be found with
dA/dt=dA/dr x dr/dt
where dA/dt is the derivative of area as a function of time. So multiplying these derivatives gives me
dA/dt=dA/dr x dr/dt
=(2∏r)(5m/s)
=(50∏rm)/s
The radius at 10 seconds by multiplying 5m/s x 10s =50m. So I substitute 50m in for r.
=(50∏(50m)m)/s
=(2500∏m2)/2

I'm struggling a bit in my calculus class and I'm a bit unsure about all of this. Am I going in the right direction with my solution attempt above?

Also I noticed on these forums that people are entering equations and formulas in "fancy text", if I may call it that. What I mean is making math stuff look like what one would see in a textbook, as opposed to using unicode characters like I did above. How can I do that?

2. Mar 28, 2013

voko

$A = \pi r^2$ and $r = vt$, so $$\frac {dA} {dt} = \frac {dA} {dr} \frac {dr} {dt} = 2 \pi r v = 2 \pi v^2 t$$

You did well till this point. But you made a mistake in arithmetic then.

To see how to write in the "fancy text" (which is properly called LaTeX), hit the Quote button on my message.

3. Mar 28, 2013

EricPowell

How did you get $r=vt$?

4. Mar 28, 2013

voko

I got that exactly the same way you got "The radius at 10 seconds by multiplying 5m/s x 10s =50m. So I substitute 50m in for r." I just used letters instead of numbers.

5. Mar 28, 2013

EricPowell

Ah of course. $v$ for velocity and $t$ for time.

I'm still a bit confused about what you did after though. What does $dr / dt$ equal?

6. Mar 28, 2013

voko

$$\frac {dr} {dt} = \frac {d(vt)} {dt} = ?$$

7. Mar 29, 2013

EricPowell

I am so confused by all of this T_T

Okay so...if I am understanding this right $dr/dt$ is the rate at which the radius changes with time. If the radius is changing at $5m/s$, does that mean that $dr/dt=r/t$?

8. Mar 29, 2013

EricPowell

Wait I think I get it.
$\frac {dr}{dt}= \frac {5m}{s}= \frac {r}{t}= \frac {vt}{t}= v$

9. Mar 29, 2013

EricPowell

Why are my fractions so tiny?

10. Mar 29, 2013

voko

If you agree that $r = vt$, then you obtain the derivative by just following the rules.

If you do not know what $r$ really is, you need to make some assumption. The assumption in this problem is that the wave propagation speed is constant, which brings you back to $r = vt$.

11. Mar 29, 2013

voko

$foo$ makes inline expressions (small). $$bar$$ makes "display" expressions (big).

12. Mar 29, 2013

EricPowell

$$\frac {dA}{dr}=2πr$$
$$\frac {dr}{dt}=v$$
$$\frac {dA}{dt}=2πr×v$$
$$=2πvt×v$$
$$=2π \frac {5m}{s}×10s× \frac {5m}{s}$$
$$=2π×50m× \frac {5m}{s}$$
$$= \frac{1570.796m^2}{s}$$

Is this correct?
Oh I messed something up there with my formatting.
Ok fixed my formatting.

Last edited: Mar 29, 2013
13. Mar 29, 2013

EricPowell

$π∏$ Just testing different pi symbols
$$π∏$$

14. Mar 29, 2013

voko

15. Mar 29, 2013

EricPowell

Thank you so much for helping me Voko :)

16. Mar 29, 2013

voko

One more thing. You can have multiline math text in the display mode:
$$line \ 1 \\ line \ 2 \\ etc.$$