1. ## Differentiation

A curve $\displaystyle y=x^2e^{-kx}$, where $\displaystyle k$ is a positive constant has stationary points of the curve are at (0,0) and (8,a).

Determine the exact values of the constants $\displaystyle k$ and $\displaystyle a$.

2. May be the first derivative at stationary point is zero.

3. Originally Posted by zzzoak May be the first derivative at stationary point is zero.
$\displaystyle \frac{dy}{dx}= x^2e^{-kx}-2xe^{-kx}=0$

cant finish

4. Product rule? Just in case a picture helps... Or, zooming in on the chain rule... ... where (key in spoiler) ...

Spoiler: ... is the product rule. Straight lines differentiate downwards (anti-diff up). ... is the chain rule. Straight continuous lines still differentiate downwards with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Spoiler: _________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

5. Originally Posted by tom@ballooncalculus Product rule? Just in case a picture helps...

Spoiler: that's what i've done. how did you get $\displaystyle k=\frac{2}{x}$ ?

6. Factorised?

7. cant see it. please show me.

8. Originally Posted by BabyMilo $\displaystyle \frac{dy}{dx}= x^2e^{-kx}-2xe^{-kx}=0$
You didn't differentiate it correctly -- that's why you can't finish off.

9. From Tom's differentiation we have [tex]f'(x) = 2xe^{-kx}-kx^2e^{-kx} = 0 \Rightarrow e^{-kx}\left(2x-kx^2\right) = 0[/Math].
Since $\displaystyle e^{-kx} \ne 0$ for any real number $\displaystyle k$, we have $\displaystyle 2x-kx^2 = 0$. We know that $\displaystyle f'(8) = 0$,
so $\displaystyle 16-64k = 0 \Rightarrow k = \frac{1}{4}$. You can find $\displaystyle a$ easily now as it's just $\displaystyle f(8)$.

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