Chapter 7

Find \(y^{\prime}\) for a. \(y=2 \cos t\) b. \(y=\cos 2 t\) c. \(y=\sin t^{2}\) d. \(y=\sin (t+\pi)\) e. \(y=\cos (\pi t-\pi / 2)\) f. \(y=(\sin t) \times(\cos t)\) g. \(\quad y=\sin ^{2}\left(t^{4}\right)\) h. \(y=\cos (\ln (t+1))\) i. \(y=\sin (\cos t)\) j. \(y=\tan \left(\frac{\pi}{2} t\right)\) k. \(y=\tan ^{2}\left(t^{2}\right) \quad\) l. \(y=\tan (\cos t)\) \(\mathrm{m} \cdot y=-\ln (\cos t)\) n. \(y=e^{\sin t}\) o. \(y=\ln (\sin t)\)

The difference quotient $$ \frac{F(t+h)-F(t)}{h} \quad \text { approximates } \quad F^{\prime}(t) $$ when \(h\) is 'small.' Make a plot of $$ y=\cos t \quad \text { and of } \quad \frac{\sin (t+0.2)-\sin t}{0.2} \quad-\frac{\pi}{2} \leq t \leq 2 \pi . $$ Repeat, using \(h=0.05\) instead of \(h=0.2\).

For small positive values of \(z, \sin z

Compute the derivative of $$ y(t)=1+2 t-5 t^{7}+2 e^{3 t}-\ln 6 t+2 \sin t-3 \cos t $$

For small positive values of \(z, z<\tan z\) (Equation 7.5\(),\) but 'just barely so' a. Compute $$ \tan z-z \quad \text { for } \quad z=0.01 $$ for \(\quad z=0.001 \quad\) and for \(\quad z=0.0001\). b. Compute $$ \frac{\tan z}{z} $$ for \(\quad z=0.01\) for \(\quad z=0.001 \quad\) and for \(\quad z=0.0001\). c. Note that the slope of the tangent to \(y=\tan t\) at (0,0) is $$ [\tan t]_{t=0}^{\prime}=\lim _{h \rightarrow 0} \frac{\tan (0+h)-\tan 0}{h}=\lim _{h \rightarrow 0} \frac{\tan h}{h} $$ What is your best estimate of $$ [\tan t]_{t=0}^{\prime} ? $$

Show that the suggested solutions solve the associated equations. a. \(y=\cos t\) \(y^{\prime \prime}+y=0\) b. \(y=\cos 2 t\) \(y^{\prime \prime}+4 y=0\) c. \(\quad y=3 \sin t+2 \cos t\) \(y^{\prime \prime}+y=0\) d. \(y=-3 \sin 2 t+5 \cos t\) \(y^{\prime \prime}+4 y=0\) e. \(y=e^{-t} \quad y^{\prime \prime}+2 y^{\prime}+y=0\) f. \(\quad y=e^{-t} \sin t\) \(y^{\prime \prime}+2 y^{\prime}+2 y=0\)

Compute the derivatives of a. \(\quad y=2 \sin t \cos t\) b. \(y=\sin ^{2} t+\cos ^{2} t\) c. \(\quad y=\sec t=\frac{1}{\cos t}\) d. \(y=\cot t\) e. \(y=\ln \cos t\) f. \(y=\sin ^{2} t-\cos ^{2} t\) g. \(\quad y=\csc t=\frac{1}{\sin t}\) h. \(y=\sec ^{2} t\) i. \(\quad y=e^{\cos t}\) j. \(\quad y=\ln (\sec t)\) k. \(\quad y=e^{-t} \sin t\) l. \(y=\tan ^{2} t\) m. \(\quad y=\frac{e^{2 t}}{400}\) n. \(y=\ln \left(\cos ^{20} t\right)\) o. \(\quad y=\frac{\ln t^{2}}{30}\)

We need the identities $$ \begin{array}{l} \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \\ \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \end{array} $$ in the next two exercises and in the next section. It is unlikely that you remember them from a trigonometry class. We hope you do remember, however, the double angle formulas 7.6 , $$ \sin (A+B)=\sin A \cos B+\cos A \sin B \quad \cos (A+B)=\cos A \cos B-\sin A \sin B $$ a. Use \(\sin (A+B)=\sin A \cos B+\cos A \sin B\) and the identities, \(\sin (-A)=-\sin A\) and \(\cos (-A)=\cos A,\) to show that $$ \sin (A-B)=\sin A \cos B-\cos A \sin B $$ b. Use the equations $$ \begin{array}{l} \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \sin (A-B)=\sin A \cos B-\cos A \sin B \end{array} $$ to show that $$ \sin (A+B)-\sin (A-B)=2 \cos A \sin B $$ c. Solve for \(A\) and \(B\) in $$ \begin{array}{l} A+B=x \\ A-B=y \end{array} $$ d. Substitute the values for \(A+B, A-B, A,\) and \(B\) into Equation 7.7 to obtain $$ \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} $$ e. Use \(\cos (A+B)=\cos A \cos B-\sin A \sin B\) to show that $$ \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) $$ The argument will be similar to the previous steps.

Show that if \(B\) and \(\omega\) are constants and \(y(t)=B \cos (\omega t),\) then $$ y(0)=B \quad y^{\prime}(0)=0 \quad \text { and } \quad y^{\prime \prime}(t)+\omega^{2} y(t)=0 . $$

Compute \(y^{\prime}\) and solve for \(t\) in \(y^{\prime}(t)=0 .\) Sketch the graphs and find the highest and the lowest points of the graphs of: a. \(y=\sin t+\cos t \quad 0 \leq t \leq \pi\) b. \(y=e^{-t} \sin t \quad 0 \leq t \leq \pi\) c. \(y=\sqrt{3} \sin t+\cos t \quad 0 \leq t \leq \pi\) d. \(y=\sin t \cos t \quad 0 \leq t \leq \pi\) e. \(y=e^{-t} \cos t \quad 0 \leq t \leq \pi\) f. \(y=e^{-\sqrt{3} t} \cos t \quad 0 \leq t \leq \pi\) Note: \(\cos ^{2} t-\sin ^{2} t=\cos (2 t)\)

Show that if \(A, B\) and \(\omega\) are constants and \(y(t)=A \sin (\omega t)+B \cos (\omega t),\) then \(y(0)=B \quad y^{\prime}(0)=\omega A\) and $$ y^{\prime \prime}(t)+\omega^{2} y(t)=0 $$

a. Use \(\cot x=\frac{\cos x}{\sin x}\) and the quotient rule to show that $$ [\cot x]^{\prime}=-\csc ^{2} x $$ b. Use \(\sec x=\frac{1}{\cos x}=(\cos x)^{-1}\) and the power chain rule to show that $$ [\sec x]^{\prime}=\sec x \tan x $$ c. Show that $$ [\csc x]^{\prime}=-\csc x \cot x . $$

At least, show that $$ y(t)=e^{-b t} \cos \left(\sqrt{c-b^{2}} t\right) $$ solves $$ y^{\prime \prime}(t)+2 b y^{\prime}(t)+c y(t)=0 \quad \text { for } \quad c-b^{2}>0 $$

The complete solution to Exercise 7.4 .7 is given in the Solutions Section on the web. Try your hand with showing that $$ y(t)=e^{-b t} \sin \left(\sqrt{c-b^{2}} t\right) $$ solves $$ y^{\prime \prime}(t)+2 b y^{\prime}(t)+c y(t)=0 \quad \text { for } \quad c-b^{2}>0 \quad \text { for } \quad b=1, \quad c=2 $$ Get a big piece of paper.

The derivative of \(y=\cos x\) is defined by $$ [\cos x]^{\prime}=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h} $$ Make a plot of $$ y=-\sin t \quad \text { and of } \quad \frac{\cos (t+0.2)-\cos t}{0.2} \quad-\frac{\pi}{2} \leq t \leq 2 \pi $$ Repeat, using \(h=0.05\) instead of \(h=0.2\).

Alternate derivation of \([\cos x]^{\prime}=-\sin x\) using implicit differentiation and \([\sin x]^{\prime}=\cos x\) Compute the derivatives of both sides of the identity $$ \sin ^{2} x+\cos ^{2} x=1 \quad \text { and obtain } \quad 2 \sin x \cos x+2 \cos x[\cos x]^{\prime}=0 $$ Use the last equation to argue that if \(\cos x \neq 0\) then $$ [\cos x]^{\prime}=-\sin x $$

How is the preference in calculus for radian measure of angles for trigonometric functions similar to the preference of the number \(e\) as the base of exponential functions?