Finding an equation of a curve?

• Jun 26th 2008, 02:43 PM
doughnuts
Finding an equation of a curve?
I've never come across a question like this before, perhaps someone will be able to point me in the right direction.
Find the equation of a curve with x-intercept equal to 2 and whose tangent line at any point (x,y) has slope $xe^{-y}$.

Watch this be something really easy I skipped over in my text book, anyways thanks in advance for the help.
• Jun 26th 2008, 03:02 PM
Moo
Hello,

Quote:

Originally Posted by doughnuts
I've never come across a question like this before, perhaps someone will be able to point me in the right direction.
Find the equation of a curve with x-intercept equal to 2 and whose tangent line at any point (x,y) has slope $xe^{-y}$.

Watch this be something really easy I skipped over in my text book, anyways thanks in advance for the help.

If the tangent line at any point has slope $xe^{-y}$, this means that $y'=xe^{-y}$, because y' represents the slope of the tanget.

So you have to solve for y in $y'=xe^{-y}$ and with $y(2)=0$, because x-intercept is 2.

$y'=xe^{-y}=\frac{x}{e^y} \implies y'e^y=x$.

Integrating both parts, we have :
$e^y=\frac{x^2}{2}+C$.

$y=\ln \left(\frac{x^2}{2}+C\right)$

Determine C thanks to the information $y(2)=0$
• Jun 26th 2008, 03:02 PM
wingless
The slope of the tangent line of the curve f(x) at the point $(x,y)$ is $\frac{dy}{dx}$.

So this question basically says $\frac{dy}{dx} = xe^{-y}$. Solve it and use the x-intercept value to find the integration constant.
• Jun 26th 2008, 03:39 PM
doughnuts
Wow that was way simpler then I thought it would be. The question was grouped in with a bunch of parametric equation questions so I thought it would be something along those lines. Thanks for the help.