Finding Equations of Normal Lines Parallel to a Given Line

SherlockOhms
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Homework Statement


Find the equation of each of the normal lines to the curve y = x^3 - 4x that is parallel to the line x + 8y - 8 = 0.


Homework Equations


Differentiation, y - y1 = m(x - x1)


The Attempt at a Solution


Well, clearly I start by differentiating y = x^3 - 4x to gey dy/dx = 3x^2 - 4. Then, this is the slope of the tangent to the graph y. This slope is equal to the slope of the line x + 8y - 8 = 0. Which is -8. So, I then evaluate for x and sub back into dy/dx, giving me the slope of the tangent. The slope of the normal line is the negative reciprocal of this. If you solve for x using
3x^2 - 4 = -8, you get an imaginary value though. Where have I gone wrong?
 
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You have the gradient of the given line wrong.
 
CAF123 said:
You have the gradient of the given line wrong.
Cool. See where I went wrong now. Stupid slip.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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