# 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:
$$line \ 1 \\ line \ 2 \\ etc.$$