MA005: Calculus I

Practice 1: $\log _{9} 20=\frac{\log (20)}{\log (9)} \approx 1.3634165 \approx \frac{\ln (20)}{\ln (9)}, \log _{3} 20=\frac{\log (20)}{\log (3)} \approx 2.726833 \approx \frac{\ln (20)}{\ln (3)}$

                   $\log _{\pi} \mathrm{e}=\frac{\log (\mathrm{e})}{\log (\pi)} \approx \mathbf{0 . 8 7 3 5 6 8 5} \approx \frac{\ln (\mathrm{e})}{\ln (\pi)}=\frac{1}{\ln (\pi)}$

Practice 2: $\mathbf{D}\left(\log _{10}(\sin (\mathrm{x}))\right)=\frac{1}{\sin (\mathrm{x}) \cdot \ln (10)} \quad \mathbf{D}(\sin (\mathrm{x}))=\frac{\cos (\mathbf{x})}{\sin (\mathrm{x}) \cdot \ln (10)}$

                 $\mathbf{D}\left(\log _{\pi}\left(\mathrm{e}^{\mathrm{x}}\right)\right)=\frac{1}{\mathrm{e}^{\mathrm{x}} \cdot \ln (\pi)} \mathrm{D}\left(\mathrm{e}^{\mathrm{x}}\right)=\frac{\mathbf{e}^{\mathbf{x}}}{\mathrm{e}^{\mathrm{x}} \cdot \ln (\pi)}=\frac{1}{\ln (\pi)}$

Practice 3: $D\left(\sin \left(2^{x}\right)\right)=\cos \left(2^{x}\right) D\left(2^{x}\right)=\cos \left(2^{x}\right) \cdot 2^{x} \cdot \ln (2)$

                 $\frac{\mathbf{d}}{\mathbf{d t}} 3^{\left(\mathrm{t}^{2}\right)}=3^{\left(\mathrm{t}^{2}\right)} \ln (3) \mathbf{D}\left(\mathrm{t}^{2}\right)=3^{\left(\mathrm{t}^{2}\right)} \ln (3) \cdot 2 \mathrm{t}$

Practice 4: $\mathrm{T}=\frac{72}{1+\mathrm{h}}=\frac{72}{1+\mathrm{t}+\sin (\mathrm{t})}$

                  $\frac{\mathbf{d} \mathrm{T}}{\mathbf{d} \mathbf{t}}=\frac{(1+\mathrm{t}+\sin (\mathrm{t})) \cdot \mathbf{D}(72)-72 \cdot \mathbf{D}(1+\mathrm{t}+\sin (\mathrm{t}))}{(1+\mathrm{t}+\sin (\mathrm{t}))^{2}}=\frac{-72(1+\cos (\mathrm{t}))}{(1+\mathrm{t}+\sin (\mathrm{t}))^{2}}$

                  When $t=5, \frac{\mathbf{d} T}{\mathbf{d t}}=\frac{-72(1+\cos (5))}{(1+\mathrm{t}+\sin (5))^{2}} \approx-\mathbf{3 . 6 3 6 9 5}$.

Practice 5: $\mathrm{x}(\mathrm{t})=3 \cos (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=-3 \sin (\mathrm{t}) \cdot \mathrm{y}(\mathrm{t})=2 \sin (\mathrm{t})$

                 so $\mathrm{dy} / \mathrm{dt}=2 \cos (\mathrm{t}) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{2 \cos (\mathrm{t})}{-3 \sin (\mathrm{t})}$.

                  When $t=\pi / 2, \frac{d y}{d x}=\frac{2 \cos (\pi / 2)}{-3 \sin (\pi / 2)}=\frac{2 \cdot 0}{-3 \cdot 1}=0 .$ (See Fig. 12)

Practice 6: When $x=1$ : pos., pos., pos. When $x=3$ : pos., neg., neg.

Practice 7: $\mathrm{x}(\mathrm{t})=3 \sin (\mathrm{t})$ and $\mathrm{y}(\mathrm{t})=3 \cos (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=3 \cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=-3 \sin (\mathrm{t}).$ Then

Practice 8: $\mathrm{x}(\mathrm{t})=3 \cos (\mathrm{t})$ and $\mathrm{y}(\mathrm{t})=2 \sin (\mathrm{t})$ so $\mathrm{dx} / \mathrm{dt}=-3 \sin (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \cos (\mathrm{t}) .$ Then