A curve has gradient e(power 4x)+e(power -x) at the point (X,Y). Given that the curve passes through the point (0,3), find the equation of the curve.

Plz Help!

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- Jun 24th 2011, 08:35 AMKhevishEquation of a curve
A curve has gradient e(power 4x)+e(power -x) at the point (X,Y). Given that the curve passes through the point (0,3), find the equation of the curve.

Plz Help! - Jun 24th 2011, 08:40 AMAckbeetRe: Equation of a curve
What ideas have you had so far?

- Jun 24th 2011, 08:45 AMKhevishRe: Equation of a curve
no idea at all!(Crying)

- Jun 24th 2011, 08:46 AMAckbeetRe: Equation of a curve
Well, you're given the derivative of a function. You'd like to get the function itself. How can you do that?

- Jun 24th 2011, 08:47 AMProve ItRe: Equation of a curve
The gradient function is $\displaystyle \displaystyle e^{4x} + e^{-x} $.

If $\displaystyle \displaystyle \frac{dy}{dx} = e^{4x} + e^{-x}$, then what is $\displaystyle \displaystyle y$? In other words, how do you undo taking a derivative? - Jun 24th 2011, 08:50 AMKhevishRe: Equation of a curve
Dunno! (Rain)

- Jun 24th 2011, 08:52 AMAckbeetRe: Equation of a curve
Have you studied integration at all?

- Jun 24th 2011, 08:55 AMKhevishRe: Equation of a curve
Yes, but did not do any of my homeworks..!!! So i forgot everything...!!

- Jun 24th 2011, 08:57 AMe^(i*pi)Re: Equation of a curve
The integration in question uses $\displaystyle \int e^{ax} = \dfrac{1}{a}e^{ax} +C$. If you don't know

*why*this is the case you'll need to use a textbook or ask your teacher for some more tuition and/or examples on integration - Jun 24th 2011, 08:58 AMAckbeetRe: Equation of a curve
Well, the most basic thing about integration is that it is, up to a constant, the inverse of differentiation. Let's say I take the function f(x) = sin(x), and I differentiate it to obtain f'(x) = cos(x). Well now, I can integrate it to get back to my original function:

$\displaystyle \int\cos(x)\,dx=\sin(x)+C.$

Notice the constant there. So use the antiderivative that e^(i*pi) just gave you, plus this fact, and then use the constant of integration to make your curve go through the desired point. Does that make sense? - Jun 24th 2011, 09:01 AMProve ItRe: Equation of a curve
- Jun 26th 2011, 12:18 PMmr fantasticRe: Equation of a curve