Thread: [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)

can somebody please explain..?

Can't see the pic but, pythag

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?

"D" represents the distance from the origin to the moving point. That's why you use the Pythagorean theorem:

I made you a picture:



Since you already know that , you can plug it in for "y" in the equation ----->

From there, take the derivative of D. As you know, the derivative of that equation is

Since you know that , you can just plug it in and multiply it by the derivative of D. That's how you're left with

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?

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).

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