# Derivative question.

1. Sep 20, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data
Find the rate of change of depth D with respect to time t
Find the rate of change of depth when the depth is 22.5 m

2. Relevant equations

D=10(2-(t/14))^2

3. The attempt at a solution
I know I need to determine the derivative for both.
dy/dx =10(2-(t/14))^2
= 10(4+(t^2/196)-2t/14)
= 40 - 10t^2/196 - 40t/14
=-20t/196-40/14
It just doesn't look right to me :/

2. Sep 20, 2009

### CompuChip

That looks fine, except that you accidentally turned a plus sign into a minus sign when you opened the brackets: 10(4 + t^2) is 40 + 10t^2, not 40 - 10t^2 :)
That also affects the derivative.

Now find out for which time, the depth D is equal to 22.5 m, and plug that number into the derivative.

(Also, your notation could use some work, but I will leave that until after you have solved the question).

3. Sep 20, 2009

### fghtffyrdmns

Oh, wait. But then the derivative is positive.

I should say the question: In order to make repairs to a retaining wall that is 40 m high, the water in a conservation reservoir is slowly drained out over a 4-week period. Starting at t=0, the water is released that after t days the water depth D (in metres) is approximated by: (the equation I gave).

Did I do something wrong?

4. Sep 20, 2009

### mathie.girl

Is this the equation you have?

$$D=10(2-\frac{t}{14})^2$$

If so, then wouldn't you use the chain rule to get:

$$\frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14})$$

If you were avoiding the chain rule, and expanding instead, then you should have a -4 instead of a -2 here:

And you also shouldn't say that it's equal to $$\frac{dy}{dx}$$, when you haven't differentiated yet. Then below, I don't think it should be a negative ten, but rather positive ten here:

Once that's fixed it should be equal to the same derivative I got with the chain rule and that will be negative if you fill in 22.5.

5. Sep 20, 2009

### fghtffyrdmns

Ahhh, thank you! I see where I made the mistake. We were supposed to use the power rule on this question.

I got my derivative as: 5t/49 -20/7

is this correct? It looks different than what I got before but I think it's 'cause I simplified it.

6. Sep 20, 2009

### fghtffyrdmns

Hmm.
Find the rate of change of depth when the depth is 22.5 m

My answer for this is -15/7. I don't think this is right though. I equated D= 25 and solved for t which gave me 7 days. So does this mean in a week the rate of change, when the water is 22.5 m, is -15/7?

7. Sep 20, 2009

### mathie.girl

When I did it, I got:

$$\frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14}) = -\frac{40}{14} + \frac{20t}{196} = \frac{5t}{49} - \frac{20}{7}$$

which is exactly what you got. So I think it's likely right. For future reference, if you put the numbers between tags like these: [tex ][/tex ] (without the spaces), you can normally make it a lot nicer looking. Between these tags, fractions are written like \frac{numerator}{denominator}, which looks like $$\frac{numerator}{denominator}$$.

Did you get -10/7 or -15/7? Not sure either is right...

Last edited: Sep 20, 2009
8. Sep 20, 2009

### fghtffyrdmns

Oh, thank you.

I just solved it like this: $$22.5 = 10(2-(t/14))^2$$

$$2.25=(2-(t/14))^2$$
took the square root of both sides

$$1.5 = 2 -t/14$$
$$t = (1/2)(14)$$
t= 7.

Do I sub this into the derivative or the function?

9. Sep 20, 2009

### mathie.girl

That looks good. Since you're looking for the rate of change, you sub it into the derivative. And yup, -15/7 looks right.

Sorry about doubting you before - I must have been having a weak moment!

10. Sep 20, 2009

### fghtffyrdmns

Thank you! I finally understand this much better :)!

11. Sep 20, 2009

### mathie.girl

No problem. Good luck with the course!