# Related Rates

• May 1st 2010, 04:53 PM
Lukybear
Related Rates
The altitude of a right-angled triangle is 6 cm and the base is increasing at a constant rate of 2cm/s. At what rate it the hypotenuse increasing when its length is 10 cm?

No idea how to do this question. But i did approach it. I tried using similar triangles but it didn't work out. Why?

Let length be y and base be x

At y=10
6/y = x/h
6/10 = x/h
x = 3/5 h
dx/dt = 3/5 dh/dt
thus dh/dt = 10/3 cm/s

The actual answer is 1.6 :(
• May 1st 2010, 05:31 PM
Zion
Since it's a right triangle you need to use the pythagorean theorem...

$x^2 + y^2 = z^2$

And since we are dealing with rates you need to derive the equation

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

The 2's cancel so we're left with

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

now what we're looking for is how fast the hypotenuse is changing, $\frac{dz}{dt}$

and what we are given is
$x = 6$

$\frac{dx}{dt} = 0$

$y=8$ Use pythagorean theorem here to get $y$

$\frac{dy}{dt} = 2$

and

$z = 10$

then we just solve

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

$0 + 8*2 = 10\frac{dz}{dt}$

$\frac{dz}{dt} = \frac{16}{10} = 1.6$

Hope this helps! it's my first time posting a reply so sorry if it's unclear, let me know if you have any questions
• May 1st 2010, 09:01 PM
Lukybear
Quote:

Originally Posted by Zion
Since it's a right triangle you need to use the pythagorean theorem...

$x^2 + y^2 = z^2$

And since we are dealing with rates you need to derive the equation

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

The 2's cancel so we're left with

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

now what we're looking for is how fast the hypotenuse is changing, $\frac{dz}{dt}$

and what we are given is
$x = 6$

$\frac{dx}{dt} = 0$

$y=8$ Use pythagorean theorem here to get $y$

$\frac{dy}{dt} = 2$

and

$z = 10$

then we just solve

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

$0 + 8*2 = 10\frac{dz}{dt}$

$\frac{dz}{dt} = \frac{16}{10} = 1.6$

Hope this helps! it's my first time posting a reply so sorry if it's unclear, let me know if you have any questions

Wow thats great. But i dont really understand this.

The altitude, the line from the vertex of right angle triangle to its perpendicular opposite is 6. And it is given dx/dt = 2 m/s

• May 1st 2010, 10:28 PM
Debsta
Quote:

Originally Posted by Lukybear
Wow thats great. But i dont really understand this.

The altitude, the line from the vertex of right angle triangle to its perpendicular opposite is 6. And it is given dx/dt = 2 m/s

Here's an easier wayI think.
First of all redraw your diagram. Draw it (I don't know how to here so I'm giving you a verbal description - sorry) with the right angle in the bottom left corner. The height (vertical line on left) is 6cm, call the base x and the hypotenuse h. Your initial diagram was incorrect.
Tell me when you are done and I'll help with the next step.
• May 1st 2010, 10:57 PM
Debsta
You still there? Happy to walk you through the steps of related rates problems if interested.
• May 2nd 2010, 04:33 AM
Lukybear
Quote:

Originally Posted by Debsta
Here's an easier wayI think.
First of all redraw your diagram. Draw it (I don't know how to here so I'm giving you a verbal description - sorry) with the right angle in the bottom left corner. The height (vertical line on left) is 6cm, call the base x and the hypotenuse h. Your initial diagram was incorrect.
Tell me when you are done and I'll help with the next step.

But if just going on my diagram how would you do this problem.

As I am pretty sure that the altitude is that line joining the vertex to perpendicular opposite.
• May 2nd 2010, 05:29 PM
Debsta
Quote:

Originally Posted by Lukybear
But if just going on my diagram how would you do this problem.

As I am pretty sure that the altitude is that line joining the vertex to perpendicular opposite.

A triangle actually has 3 altitudes, depending on which side you define to be the base - altitude and "base" are perpendicular. Your diagram is incorrect for this problem.
• May 2nd 2010, 11:53 PM
Lukybear
Quote:

Originally Posted by Debsta
A triangle actually has 3 altitudes, depending on which side you define to be the base - altitude and "base" are perpendicular. Your diagram is incorrect for this problem.

So your saying that the hypotenuse is between the base and the altitude?
• May 3rd 2010, 04:18 AM
Debsta
Quote:

Originally Posted by Lukybear
So your saying that the hypotenuse is between the base and the altitude?

yes