# speed derivative

Printable View

• Jul 10th 2007, 11:20 AM
Samantha
speed derivative
A point moves along the curve $\displaystyle y=2x^2 + 1$ in such a way that the y value decreasing at the rate of 2 units per second. At what rate is x chaning when $\displaystyle x = \frac {3} {2}$?
• Jul 10th 2007, 11:37 AM
CaptainBlack
Quote:

Originally Posted by Samantha
A point moves along the curve $\displaystyle y=2x^2 + 1$ in such a way that the y value decreasing at the rate of 2 units per second. At what rate is x chaning when $\displaystyle x = \frac {3} {2}$?

If $\displaystyle y=2x^2 + 1$ then differentiate with respect to t:

$\displaystyle \frac{dy}{dt}=2x \frac{dx}{dt}$

and the rest is easy.

RonL
• Jul 10th 2007, 11:40 AM
Samantha
decreasing $\displaystyle \frac {1} {3} unit/sec$?
• Jul 10th 2007, 11:46 AM
Soroban
Hello, Samantha!

Quote:

A point moves along the curve: $\displaystyle y \:=\:2x^2 + 1$ in such a way
that the y value decreasing at the rate of 2 units/sec.
At what rate is $\displaystyle x$ changing when $\displaystyle x = \frac {3} {2}$ ?

Differentiate with respect to time: .$\displaystyle \frac{dy}{dt} \:=\:4x\cdot\frac{dx}{dt} \quad\Rightarrow\quad \frac{dx}{dt}\;=\;\frac{1}{4x}\cdot\frac{dy}{dt}$

We are given: .$\displaystyle \frac{dy}{dt} = -2,\;x = \frac{3}{2}$

Therefore: .$\displaystyle \frac{dx}{dt} \;=\;\frac{1}{4\left(\frac{3}{2}\right)}(-2) \;=\;-\frac{1}{3}$ units/sec