1. ## Another Geographic Derivative

I'm stuck on this question:

In the curve to which the equation is $\displaystyle y^2+x^2=4$, find the value of $\displaystyle x$ at those points where the slope = 1.
I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for $\displaystyle y$, and then take the derivative?

$\displaystyle y^2+x@=4$

$\displaystyle y^2=4-x^2$

$\displaystyle y=\pm\sqrt{4-x^2}$

is this right so far? Thanks.

2. Originally Posted by dbakeg00
I'm stuck on this question:

I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for $\displaystyle y$, and then take the derivative?

$\displaystyle y^2+x@=4$

$\displaystyle y^2=4-x^2$

$\displaystyle y=\pm\sqrt{4-x^2}$

is this right so far? Thanks.
You can use implicit differentiation to find $\displaystyle \frac{dy}{dx}$

$\displaystyle 2y\frac{dy}{dx}+2x=0$

$\displaystyle \frac{dy}{dy}=-\frac{x}{y}$

Set the derivative equal to 1

$\displaystyle -\frac{x}{y}=1$

$\displaystyle y=-x$

Substitute back into the original equation and solve for x

3. Thank you...guess I haven't learned howto do that yet.

4. Originally Posted by dbakeg00
I'm stuck on this question:

I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for $\displaystyle y$, and then take the derivative?

$\displaystyle y^2+x@=4$

$\displaystyle y^2=4-x^2$

$\displaystyle y=\pm\sqrt{4-x^2}$

is this right so far? Thanks.
Yes. Find the derivative, set it equal to 1 and solve for x

5. Originally Posted by dbakeg00
I'm stuck on this question:

I am not looking for the entire answer, just where to start. Would the first thing to do be to solve for $\displaystyle y$, and then take the derivative?

$\displaystyle y^2+x@=4$

$\displaystyle y^2=4-x^2$

$\displaystyle y=\pm\sqrt{4-x^2}$

is this right so far? Thanks.
Yes. Just find the derivative, set it equal to 1 and solve for x

6. I dont even know how to go about setting that equal to 1 and solving...any tips on how to start?

7. Originally Posted by dbakeg00
I dont even know how to go about setting that equal to 1 and solving...any tips on how to start?
Break y up into 2 functions

$\displaystyle y=\sqrt{4-x^2}$ and $\displaystyle y=-\sqrt{4-x^2}$

For $\displaystyle y=\sqrt{4-x^2}=(4-x^2)^{\frac{1}{2}}$

$\displaystyle y'=\frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)$

Set it equal to 1 and solve for x

$\displaystyle \frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)=1$

$\displaystyle \frac{-x}{\sqrt{4-x^2}}=1$

[tex]-x=\sqrt{4-x^2}[/math

$\displaystyle x^2=4-x^2$

Finish it up

8. got it solved. thanks for your help!