# Math Help - [SOLVED] Related Rates?

1. ## [SOLVED] Related Rates?

Find the rate of change of the distance between the origin and a moving point on the graph of y = x² + 1 if dx/dt = 2 cm/sec.

Here's the solution:
http://calcchat.tdlc.com/solutionart/calc8e/02/f/se02f01013.gif
except I don't understand where they got that first equation D = √(x² + y²)

2. Can't see the pic but, pythag

3. Originally Posted by tom@ballooncalculus
Can't see the pic but, pythag
But why would the pythagorean theorem apply here though..? I don't see a triangle in the graph..
here's a working link to the solution if you want to look:
http://i134.photobucket.com/albums/q...13solution.jpg

Also, how would I do this problem if instead of y = x² + 1, it's y = sin x?

4. "D" represents the distance from the origin to the moving point. That's why you use the Pythagorean theorem: $D=\sqrt{x^2 +y^2}$

Since you already know that $y=x^2+1$, you can plug it in for "y" in the equation $D=\sqrt{x^2 +y^2}$ -----> $D=\sqrt{x^2 +(x^2+1)^2}$

From there, take the derivative of D. As you know, the derivative of that equation is $\frac{dD}{dt}=\frac{1}{2} (x^4+3x^2+1)^{\frac{-1}{2}}(4x^3+6x)\frac{dx}{dt}$

Since you know that $\frac{dx}{dt} = 2$, you can just plug it in and multiply it by the derivative of D. That's how you're left with $\frac{4x^3+6x}{\sqrt{x^4+3x^2+1}}$

5. So when they say origin, they mean the vertex of the graph? ..or does it even matter where you draw the point at?
Like if in the same problem, y = sin(x) instead of y = x² + 1, would I draw a triangle the same way you did, with the point at (0, 0) instead, and then use the pythagorean theorem? or does it require a different method because it's a trig function?

6. Ooops. I don't know why I did that. o_O :P

It's supposed to be at (0,0), the origin. Here's a fixed graph:

And yes, you would still draw a triangle from the origin in that case. Remember, you're finding the distance from the origin to a point (x,y).

7. haha, ok. I understand it now. Thank you! Your graph really helped!