Finding the tangent line to $y= 7 \sin( \pi x+y)$ at the point $(-1,0)$

Question

Find the lines that are tangent and normal to the curve at the given point $$y= 7 \sin( \pi x+y), \qquad (-1,0)$$ The line tangent to the curve $y= 7 \sin (\pi x+y)$ at $(-1,0)$ is $y=\ ?$

How would I solve this?

Answer 1

Hint: Use Implicit Differentiation. We get $$\frac{dy}{dx}=7\left(\pi+\frac{dy}{dx}\right)\cos(\pi x+y).$$

Added: Now we could solve for $\frac{dy}{dx}$ and substitute. I prefer to substitute and then solve. So let $m$ be the value of $\frac{dy}{dx}$ at $(-1,0)$. Substituting, we obtain $$m=7(\pi +m)(-1).$$ Solve for $m$. We get $m=-\frac{7\pi}{8}$.

This $m$ is the slope of our tangent line. Now I expect you can find the equation of the line that passes through $(-1,0)$ and has slope $-\frac{7\pi}{8}$.

For the equation of the normal, note that the slope of the normal is $\frac{8}{7\pi}$.

Answer 2

Differentiating implicitly: $$y'=7\cos (\pi x+y)(\pi +y')\Rightarrow y'=7\pi \cos(\pi x+y)+7\cos (\pi x+y)y' $$ $$\Rightarrow y'=\frac{7\pi \cos (\pi x+y)}{1-7\cos (\pi x+y)} .$$

The slope of the tangent line at $P(-1,0)$ is $$m=y'(-1,0)=\frac{7\pi \cos (-\pi)}{1-7\cos (-\pi)}=-\frac{7\pi}{8}. $$

Then the tangent line to the curve given in $P=(x_p,y_p)=(-1,0)$ is $$y-y_p =m(x-x_p),$$ i.e., $$y-0=-\frac{7\pi}{8} (x+1)\Rightarrow y=\frac{7\pi}{8}x+ \frac{7\pi}{8},$$

and how the slope of the normal line is $$m_n=-\frac{1}{m}=\frac{8}{7\pi} $$

follows that the normal curve in $P$ is

$$y-y_p=m_n(x-x_p)\Rightarrow y=\frac{8}{7\pi}(x+1). $$