Find the series of equations which are perpendicular to y=2x^2?

Question

anonymous · Answer

I found the slope -1/4x

tkhunny · Answer

Excellent.  Now, you need an antiderivative of that.

anonymous · Answer

That just means integrate it, right?

anonymous · Answer

@tkhunny

tkhunny · Answer

Mostly.  Let's see what you get.

anonymous · Answer

$$-\frac{ x^{2} }{ 8 } +c$$

tkhunny · Answer

??  Let's clarify.  You had $y = 2x^{2}$.  This gives $\dfrac{dy}{dx} = 4x$.

Therefore, the curevs we seek have $\dfrac{dy}{dx} = -\dfrac{1}{4x}$, resulting in what antiderivative?

anonymous · Answer

I think I'm a little confused... I figured the antiderivative of -1/4x would be what I said

tkhunny · Answer

Better brush up on your exponents.  Negative One (-1) is special!

$\int -\dfrac{1}{4x}\;dx = -\dfrac{1}{4}\int \dfrac{1}{x}\;dx = -\dfrac{1}{4}\cdot\ln(|x|) + C$

anonymous · Answer

I do that every time! Would that be my equation? In my notes they have a y added and I'm not sure why