Implicit differentiation: tangent line equation

Question

I need just a little bit of help. I feel like I'm making a simple mistake. Given:

$$\sqrt{x^2+y^2}=\tan{[\frac{\pi}{4}(x+y)]}$$

and $P(0,1)$. I need to find the equation of the tangent line at point $P$.

I differentiated the entire equation, and y'(x) came out to be a ridiculous fraction, which in the end, when I plugged in x=0, I received a denominator of 0 because there was a 2x in the denominator, deeming all of my efforts worthless.

Is there a simpler way to do this than differentiating every single term, am I missing something? How would I start? Thank you for any help!

Answer 1

If we regard $y$ as a function of $x$ in some neighborhood around $x_0=0$ and derive both sides we get $$\left(x^2+y^2\right)^{-1/2}(x+y')=\sec^2\left[\frac{\pi}{4}\left(x+y\right)\right]\cdot\frac{\pi}{4}(1+y')...(*)$$ In order to find the slope of tangent line at $P$ we need $y'(x_0)=y'(0)$, so we set $x=0$ and $y=1$ in $(*)$: \begin{align*} (0+1)^{-1/2}\left[0+y'(0)\right]&=\sec^2\left[\frac{\pi}{4}\left(0+1\right)\right]\cdot\frac{\pi}{4}\left[1+y'(0)\right] \\ y'(0)&=\frac{\pi}{4}\sec^2\left(\frac{\pi}{4}\right)\left[1+y'(0)\right]\\ y'(0)&=\frac{\pi}{2}+\frac{\pi}{2}y'(0)\\ y'(0)&=-\frac{\pi}{\pi-2} \end{align*}

Answer 2

Implicitly:

$$\frac{x\,dx+y\,dy}{\sqrt{x^2+y^2}}=\frac \pi4\left(\frac{dx+dy}{\cos^2\left[\frac\pi4(x+y)\right]}\right)\iff$$

$$\left(\frac y{\sqrt{x^2+y^2}}-\frac\pi{4\cos^2\left[\frac\pi4(x+y)\right]}\right)dy=\left(-\frac x{\sqrt{x^2+y^2}}+\frac\pi{4\cos^2\left[\frac\pi4(x+y)\right]}\right)dx$$

And from here, on $\;(0,1)$ :

$$\left.\frac{dy}{dx}\right|_{(0,1)}=\frac{\left(-0+\frac\pi{4\cos^2\frac\pi4}\right)}{1-\frac\pi{4\cos^2\frac\pi4}}=\frac\pi{2-\pi}$$

Check the above: I should be sleeping already.

Answer 3

i followed a similar method, but used the abbreviation $$ \xi = \frac{\pi}4\sqrt{x^2+y^2}\sec^2\left(\frac{\pi}4(x+y)\right) $$ so obtained $$ \frac{ydy+xdx}{dy+dx} = \xi $$ from which $$ \frac{dy}{dx} = \frac{\xi-x}{y-\xi} $$ at (0,1) we find $\xi=\frac{\pi}2$ hence $$ \frac{dy}{dx}|_{(0,1)}=\frac{\pi}{2-\pi} $$