# Thread: Find the Rate of Change

1. ## Find the Rate of Change

Find the rate of change of the distance between the origin and a moving point on the graph of
y = x^2 + 1 It dx/Dr = 2 cm/second

My work:

I proceeded taking the implicit derivative of the given function to find dy/dt.

dy/dt = 2x* dx/dt

dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

dy/dt = 2(0)* (0)

2(2x^3+3x)/sqrt(x^4+3x^2+1)

My Answer is zero and the book's answer is a crazy rational function.

2. ## Re: Find the Rate of Change

Originally Posted by nycmath
Find the rate of change of the distance between the origin and a moving point on the graph of
y = x^2 + 1 It dx/Dr = 2 cm/second

My work:

I proceeded taking the implicit derivative of the given function to find dy/dt.

dy/dt = 2x* dx/dt

dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

dy/dt = 2(0)* (0)

2(2x^3+3x)/sqrt(x^4+3x^2+1)

My Answer is zero and the book's answer is a crazy rational function.

-Dan

3. ## Re: Find the Rate of Change

Originally Posted by nycmath
Find the rate of change of the distance between the origin and a moving point on the graph of
y = x^2 + 1 It dx/Dr = 2 cm/second

My work:

I proceeded taking the implicit derivative of the given function to find dy/dt.

dy/dt = 2x* dx/dt

dy/dt = 2x* (2) and the I plugged x = 0 from the origin value for x.

dy/dt = 2(0)* (0)

2(2x^3+3x)/sqrt(x^4+3x^2+1)

My Answer is zero and the book's answer is a crazy rational function.

Sorry, just a little joke.

The problem is asking about how fast the distance is changing between a point on the graph and the origin. So say we have a point on the graph (x, y). What is the distance between that point and the origin? Once you get that you can take your derivative. (And recall that y = x^2 - 1 before you differentiate.)

-Dan

4. ## Re: Find the Rate of Change

This is what I did:

(1) I used the distance formula for points to find d.

d = sqrt{(x-0)^2 + (x^2-1-0)}

d = sqrt{x^4-2x^2+1}

(2) I then differentiated implicitly and found dr/dt to be

[2x(2x^2-1)]/sqrt{x^4-2x^2+1}

It is not correct.

5. ## Re: Find the Rate of Change

Originally Posted by nycmath
d = sqrt{(x-0)^2 + (x^2-1-0)}
Your parabola is $\displaystyle y = x^2 + 1$. You have a minus sign there. And don't forget the (x - 0)^2 term.

-Dan

6. ## Re: Find the Rate of Change

Almost there...take a look.

dr/dt = 2(2x^3+ 3)/sqrt(x^4+3x^2+1)

The numerator should be
2(2x^3+3x). I am missing an "x" for the term 3x. See it?

7. ## Re: Find the Rate of Change

Originally Posted by nycmath
Almost there...take a look.

dr/dt = 2(2x^3+ 3)/sqrt(x^4+3x^2+1)

The numerator should be
2(2x^3+3x). I am missing an "x" for the term 3x. See it?
$\displaystyle r = \sqrt{x^4 + 3x^2 + 1}$

$\displaystyle \frac{dr}{dt} = \frac{1}{2} (x^4 + 3x^2 + 1)^{-1/2} \cdot (4x^3 + 6x) \frac{dx}{dt}$

-Dan

8. ## Re: Find the Rate of Change

I clearly see my error. When taking the derivative of the inside using the chain rule, I mistakenly multiplied 4x^3 and 6x by the value given for dx/dt or 2. Thank you for your help. Math is fun.