issues with the algebra in a problem

I have figured out how to solve this problem but I'm not getting the correct answer.

A 1000 L tank loses water so that, after t days, the remaining volume is

v(t) = 1000[1-(t/10)]^2 for 0<t<10.

How rapidly is the water being lost when the tank is half full?

I got the derivative fine. -200(1-(t/10))

I know to solve v(t) = 500 then take the value of t and input it into v'(t)

when solving v(t) = 500 I end up with the answer 10-5sqrt(2) or -5(sqrt(2) - 2)

but to get the correct answer to the question it needs to be 5sqrt2

the correct answer is 100sqrt(2) litres/day

thank you for your help!

Re: issues with the algebra in a problem

Quote:

Originally Posted by

**Jonroberts74** I have figured out how to solve this problem but I'm not getting the correct answer.

A 1000 L tank loses water so that, after t days, the remaining volume is

v(t) = 1000[1-(t/10)]^2 for 0<t<10.

How rapidly is the water being lost when the tank is half full?

I got the derivative fine. -200(1-(t/10))

I know to solve v(t) = 500 then take the value of t and input it into v'(t)

when solving v(t) = 500 I end up with the answer 10-5sqrt(2) or -5(sqrt(2) - 2)

but to get the correct answer to the question it needs to be 5sqrt2

the correct answer is 100sqrt(2) litres/day

thank you for your help!

$\displaystyle \displaystyle \begin{align*} 500 &= 1000 \left( 1 - \frac{t}{10} \right) ^2 \\ \frac{1}{2} &= \left( 1 - \frac{t}{10} \right) ^2 \\ \pm \sqrt{ \frac{1}{2} } &= 1 - \frac{t}{10} \\ \pm \frac{\sqrt{2}}{2} &= 1 - \frac{t}{10} \\ \frac{t}{10} &= 1 \pm \frac{\sqrt{2}}{2} \\ \frac{t}{10} &= \frac{2 \pm \sqrt{2}}{2} \\ t &= 5 \left( 2 \pm \sqrt{2} \right) \\ t &= 10 \pm 5\sqrt{2} \end{align*}$

Obviously since $\displaystyle \displaystyle \begin{align*} 0 \leq t \leq 10 \end{align*}$ we must accept $\displaystyle \displaystyle \begin{align*} t = 10 - 5\sqrt{2} \end{align*}$, so I agree with your value of t.

You should double check your derivative, as it should be $\displaystyle \displaystyle \begin{align*} v'(t) = -2000 \left( 1 - \frac{t}{10} \right) \end{align*}$, not $\displaystyle \displaystyle \begin{align*} -200 \left( 1 - \frac{t}{10} \right) \end{align*}$.

Re: issues with the algebra in a problem

ah okay, thank you! I thought the derivative inside the parenthesis was (-1/10) making for

1000(2)[1-(t/10)](-1/10) -> 2000[1-(1 - (t/10)](-1/10) -> -200[1 - (t/10)]

Re: issues with the algebra in a problem

Oops, you are right, I misread it >_< disregard my last response...

Re: issues with the algebra in a problem

haha no worries. if you could, would you show me how you would work through v'(t) for t = 10-5sqrt2 because I am not getting the correct answer of 100sqrt2

Thank you!

Re: issues with the algebra in a problem

Well let's see...

$\displaystyle \displaystyle \begin{align*} v' \left( 10 - 5\sqrt{2} \right) &= -200 \left( 1 - \frac{10 - 5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{10}{10} - \frac{10 - 5\sqrt{2}}{10} \right) \\ &= -200 \left[ \frac{10 - \left( 10 - 5\sqrt{2} \right) }{10} \right] \\ &= -200 \left( \frac{10 - 10 + 5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{5\sqrt{2}}{10} \right) \\ &= -200 \left( \frac{\sqrt{2}}{2} \right) \\ &= -100\sqrt{2} \end{align*}$

And of course the negative value represents the fact that the water level is decreasing...

Re: issues with the algebra in a problem

ah I see where I messed up!

Thank you!!