Question

Check my answer - related rates problem

Answer 1

The base of a right triangle is increasing at a rate of 4cm/sec while the height is increasing at the rate of 3 cm/sec. When the base is 5cm and the height is 12cm, find the rate of the perimeter.

Answer 2

I set up the equation: \[P = b + h + \sqrt{b^2 + h^2}\]
Then I took the derivative:
\[\frac{ dP }{ dt } = \frac{ db }{ dt } + \frac{ dh }{ dt } + \frac{ 1 }{ 2 }(b^2+h^2)^\frac{ -1 }{ 2 }*2b*\frac{ db }{ dt }*2h*\frac{ dh }{ dt }\]

Answer 3

Then I plugged in:
\[\frac{ dh }{ dt }=3, \frac{ db }{ dt }=4, b = 5, h = 12\]

Answer 4

I got 8 11/13 as my answer... I'm not sure if it's right...

Answer 5

I just noticed the height is supposed to be decreasing. I plugged in the numbers again using -3 as the rate dh/dt and ended up with 58 3/5 as my answer. That seems a little high.

Answer 6

Suggestion: always include units of measurement when you do a math or physics problem such as this one.

Answer 7

You want to know the rate at which the perimeter of the triangle is increasing.
Letting P=perimeter, you want\[\frac{ dP }{ dt }.\]

Answer 8

Your\[P = b + h + \sqrt{b^2 + h^2}\]

Answer 9

looks fine to me.
I'm now going to find the derivative of this expression with respect to time, t, and compare my result with yours.

Answer 10

Incorrect:\[\frac{ dP }{ dt } = \frac{ db }{ dt } + \frac{ dh }{ dt } + \frac{ 1 }{ 2 }(b^2+h^2)^\frac{ -1 }{ 2 }*2b*\frac{ db }{ dt }*2h*\frac{ dh }{ dt }\]

Answer 11

Correct:

Answer 12

\[\frac{ dP }{ dt } = \frac{ db }{ dt } + \frac{ dh }{ dt } + \frac{ 1 }{ 2 }(b^2+h^2)^\frac{ -1 }{ 2 }*[2b*\frac{ db }{ dt }+2h*\frac{ dh }{ dt }]\]

Answer 13

Think about this and see whether you agree or disagree.

Answer 14

Thanks for replying! How is it any different if there are brackets around the second part? Shouldn't it not matter since it's all multiplication?

Answer 15

Abbles, it is NOT all multiplication. Look how I replaced one of your ' * ' with a " + "

Answer 16

Ah, I see! I'm getting -3/13 when I plug in, does that sound right? Thanks for the help!