Story Problem with Derivatives

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November 18th 2009, 01:03 PM
seuzy13

Story Problem with Derivatives

Water is flowing into a tank at the following rate: $E(t) = \frac{t+2}{t^2+1}$ . The water is leaving the tank at the rate: $D(t) = sint \cdot \frac{10t}{\sqrt{t^4 + 5}}$ .

(a) What is the rate of change in the amount of water, A(t), in the tank at time t?
(b) At t = 2 is the amount of water in the tank increasing or decreasing?
(c) At t = 2 is the rate of change of the amount of water in the tank increasing or decreasing?

As for my work so far:

If I'm thinking correctly, the answer to (a) would be:
$A(t) = \frac{t + 2}{t^2 + 1} - sint \cdot \frac{10t}{\sqrt{t^4 + 5}}$ .
But I'm not sure at all.

If what I have is right however, the answer to (b) would be increasing if A(t) is positive at t =2 and decreasing if A(t) is negative at t = 2.

And for (c) I would apply the same logic as on (b) by taking the derivative of A(t) and seeing if it is positive or negative at t = 2.

The problem is that I'm very suspicious of my answers and my calculator isn't agreeing with me. Taking the derivative of A(t) is easy to mess up, and my teacher has always been implying that there shouldn't be too much writing on these types of problems that he gives us. As it is, I'm doing a lot of writing.

Help me out? Thanks!
November 18th 2009, 02:00 PM
skeeter

Quote:

Originally Posted by seuzy13

Water is flowing into a tank at the following rate: $E(t) = \frac{t+2}{t^2+1}$ . The water is leaving the tank at the rate: $D(t) = sint \cdot \frac{10t}{\sqrt{t^4 + 5}}$ .

(a) What is the rate of change in the amount of water, A(t), in the tank at time t?
(b) At t = 2 is the amount of water in the tank increasing or decreasing?
(c) At t = 2 is the rate of change of the amount of water in the tank increasing or decreasing?

As for my work so far:

If I'm thinking correctly, the answer to (a) would be:
$A(t) = \frac{t + 2}{t^2 + 1} - sint \cdot \frac{10t}{\sqrt{t^4 + 5}}$ .
But I'm not sure at all.

your expression for A(t) is correct.

If what I have is right however, the answer to (b) would be increasing if A(t) is positive at t =2 and decreasing if A(t) is negative at t = 2.

correct

And for (c) I would apply the same logic as on (b) by taking the derivative of A(t) and seeing if it is positive or negative at t = 2.

correct again.

The problem is that I'm very suspicious of my answers and my calculator isn't agreeing with me. Taking the derivative of A(t) is easy to mess up, and my teacher has always been implying that there shouldn't be too much writing on these types of problems that he gives us. As it is, I'm doing a lot of writing.

Help me out? Thanks!

you're right when you say that A'(t) is a mess. Can't guarantee it, but I get ...

$A'(t) = 10\left(\frac{(t^4-5)\sin{t}}{(t^4+5)^{\frac{3}{2}}} - \frac{t\cos{t}}{\sqrt{t^4+5}} \right) - \frac{t^2+4t-1}{(t^2+1)^2}$
November 18th 2009, 04:32 PM
seuzy13

Hm...I'm getting quite different answers every time I take a crack at it. At least we agree on what the derivative of $\frac{t + 2}{t^2 + 1}$ is.

And using trace X = 2 on the derivative in my calculator is -0.476. None of my answers or yours is coming out to that, although I obviously could have made a mistake with the calculator as well. =(
November 18th 2009, 05:19 PM
skeeter

Quote:

Originally Posted by seuzy13

Hm...I'm getting quite different answers every time I take a crack at it. At least we agree on what the derivative of $\frac{t + 2}{t^2 + 1}$ is.

And using trace X = 2 on the derivative in my calculator is -0.476. None of my answers or yours is coming out to that, although I obviously could have made a mistake with the calculator as well. =(

change your calculator to radian mode. (Wink)
November 18th 2009, 05:20 PM
seuzy13

Ah! You're right! I had it on degrees!!!
(Surprised)