Help with Inverse Trig Differentiation Question (1 Viewer)

siggy · Jan 31, 2015

Find, in the form y = mx + b, the equation of the tangent and the normal to the curve at the point indicated:

y = inverse sin(x/2), at x = sqroot(2)

I found the tangent to be y = x/root(2) - 1 but the answer has the same but with + pi/4.
I have no idea where pi came from essentially
Thx all

InteGrand · Jan 31, 2015

$y=\sin ^{-1}\frac{x}{2}$ , point in question is $\left (\sqrt{2}, \frac{\pi}{4}\right )$ (since $\sin ^{-1}\frac{\sqrt{2}}{2}=\frac{\pi}{4}$ )

$y^\prime = \frac{1}{\sqrt{4-x^2}}$

So at the given x-value, $y^\prime = \frac{1}{\sqrt{4-(\sqrt{2})^2}}=\frac{1}{\sqrt{2}}$ .

Sub. in the point $\left (\sqrt{2}, \frac{\pi}{4}\right )$ to find b in y = mx + b, where $m = \frac{1}{\sqrt{2}}$ .

$\frac{\pi}{4} = \frac{\sqrt{2}}{\sqrt{2}}+b\Rightarrow b = \frac{\pi}{4}-1$ .

siggy · Jan 31, 2015

Sweet I understand now, thanks!

Help with Inverse Trig Differentiation Question (1 Viewer)

siggy

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InteGrand

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