Chapter 1.4, Problem 18E

To determine

To Explain: If the curve has tangent at the given point.

Answer to Problem 18E

Yes, the tangent at the given point exists and the slope is $-\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information:

The curve is: $f\left(x\right)=\left\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}\right.$ and the point is $x=\frac{3\pi }{4}$ .

Concept used:

If the curve has a tangent at any point say $x=a$, then the two-sided limit of the difference quotient should exist, that is:

  $\underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

For any function f , slope at any point $x=a$ can be calculated as: $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

First check if the two-sided quotient limit of $f\left(x\right)=\left\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\\ \cos x,\frac{3\pi }{4}\le x\le 2\pi \end{array}\right.$ at $x=\frac{3\pi }{4}$ exist or not.

Let $x=h+\frac{3\pi }{4},h=x-\frac{3\pi }{4}$ .

  $\underset{h\to 0^{-}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}f\left(x\right)=\left\{\begin{array}{l}\sin x,0\le x<\frac{3\pi }{4}\hfill \\ \cos x,\frac{3\pi }{4}<x\le 2\pi \hfill \end{array}\right.$

  $=\underset{h\to 0^{-}}{\lim }\frac{\sin \left(h+\frac{3\pi }{4}\right)-\sin \left(\frac{3\pi }{4}\right)}{h}$

  $\text{Since }\sin \left(a+b\right)=\sin a\cos b+\sin b\cos a$

  $=\underset{h\to 0^{-}}{\lim }\frac{\sin \left(h\right)\cos \left(\frac{3\pi }{4}\right)+\cos \left(h\right)\sin \left(\frac{3\pi }{4}\right)-\sin \left(\frac{3\pi }{4}\right)}{h}$

  $\sin \left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}},\sin \left(0\right)=0,\cos \left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}},\text{}\cos \left(0\right)=1$

  $=\underset{h\to 0^{-}}{\lim }\frac{-\frac{1}{\sqrt{2}}\sin \left(h\right)+\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}}{h}$

  $=\underset{h\to 0^{-}}{\lim }\frac{-\frac{1}{\sqrt{2}}\sin \left(h\right)}{h}+\underset{h\to 0^{-}}{\lim }\frac{\frac{1}{\sqrt{2}}\left(\cos \left(h\right)-1\right)}{h}$

  $=-\frac{1}{\sqrt{2}}\underset{h\to 0^{-}}{\lim }\frac{\sin \left(h\right)}{h}+\frac{1}{\sqrt{2}}\underset{h\to 0^{-}}{\lim }\frac{\left(\cos \left(h\right)-1\right)}{h}\left(\text{Apply L'hopital rule}\right)$

  $=-\frac{1}{\sqrt{2}}+0$

  $=-\frac{1}{\sqrt{2}}$

Now,

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{\cos \left(h+\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)}{h}\\ \text{Since }\cos \left(a+b\right)=\cos a\cos b-\sin a\sin b\\ =\underset{h\to 0^{+}}{\lim }\frac{\cos \left(h\right)\cos \left(\frac{3\pi }{4}\right)-\sin \left(\frac{3\pi }{4}\right)\sin \left(h\right)-\cos \left(\frac{3\pi }{4}\right)}{h}\\ \sin \left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}},\cos \left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}\\ =\underset{h\to 0^{+}}{\lim }\frac{-\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}\sin \left(h\right)+\frac{1}{\sqrt{2}}}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{-\frac{1}{\sqrt{2}}\cos \left(h\right)+\frac{1}{\sqrt{2}}}{h}-\frac{1}{\sqrt{2}}\underset{h\to 0^{+}}{\lim }\frac{-\frac{1}{\sqrt{2}}\sin \left(h\right)}{h}\left(\text{Apply L'hopital rule}\right)\\ =-\frac{1}{\sqrt{2}}\end{array}$

Here it can be seen that $\underset{h\to 0^{-}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}$ . Hence the tangent at $x=\frac{3\pi }{4}$ exist.

The slope at $x=\frac{3\pi }{4}$ will be $\underset{h\to 0}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}$ .

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{f\left(\frac{3\pi }{4}+h\right)-f\left(\frac{3\pi }{4}\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos \left(h+\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)}{h}\\ \text{Since }\cos \left(a+b\right)=\cos a\cos b-\sin a\sin b\\ =\underset{h\to 0}{\lim }\frac{\cos \left(h\right)\cos \left(\frac{3\pi }{4}\right)-\sin \left(\frac{3\pi }{4}\right)\sin \left(h\right)-\cos \left(\frac{3\pi }{4}\right)}{h}\\ \sin \left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}},\cos \left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}\\ =\underset{h\to 0}{\lim }\frac{-\frac{1}{\sqrt{2}}\cos \left(h\right)-\frac{1}{\sqrt{2}}\sin \left(h\right)+\frac{1}{\sqrt{2}}}{h}\\ =\underset{h\to 0}{\lim }\frac{-\frac{1}{\sqrt{2}}\cos \left(h\right)+\frac{1}{\sqrt{2}}}{h}-\frac{1}{\sqrt{2}}\underset{h\to 0^{+}}{\lim }\frac{-\frac{1}{\sqrt{2}}\sin \left(h\right)}{h}\left(\text{Apply L'hopital rule}\right)\\ =-\frac{1}{\sqrt{2}}\end{array}$

The slope is $-\frac{1}{\sqrt{2}}$ .