# Math Help - Related Rates

1. ## 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

2. 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

3. 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

Please do help me again!

4. 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

Please do help me again!
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.

5. You still there? Happy to walk you through the steps of related rates problems if interested.

6. 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.

7. 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.

8. 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?

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