Chapter 5: Problem 47
\(|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}\) \(|\mathbf{x}|-\mathrm{a}>1 \quad\) for no solution \(|x|>1+a\)
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\(\mathrm{A}=\frac{1}{2} \mathrm{x}^{2} \sin \theta\) $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{x} \sin \theta \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{x}^{2}}{2} \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}$ $\quad=12 \times \frac{1}{\sqrt{2}} \times \frac{1}{12}+\frac{144}{2} \times \frac{1}{\sqrt{2}} \times \frac{\pi}{180}$ \(=2^{1 / 2}\left(\frac{1}{2}+\frac{\pi}{5}\right)\)
\(\mathrm{h}=\mathrm{ar}+\mathrm{b}\) $\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{a} \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \mathrm{a}=3, \mathrm{~b}=3$ \(\mathrm{v}=3 \pi \mathrm{r}^{2}(\mathrm{r}+1)\) $\frac{\mathrm{dv}}{\mathrm{dt}}=3 \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}+2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$ \(1=3 \pi\left(3-6^{2}+12\right) \frac{\mathrm{dr}}{\mathrm{dt}}\) \(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{360 \pi}\) Now, when \(\mathrm{r}=36\) $\frac{d v}{d t}=3 \pi\left(3 \cdot(36)^{2}+72\right) \times \frac{1}{360 \pi}$ \(=33\)
\(y=x^{3}+a x\) \(\frac{d y}{d x}=3 x^{2}+a\) \(\frac{d y}{d x}=3+a\) \(y=b x^{2}+c\) \(\frac{d y}{d x}=2 b x=-2 b\) \(\Rightarrow 3+a=-2 b\) \(\Rightarrow 3+a+2 b=0\) As \(-1-a=0\) \(\Rightarrow a=-1\) \(b+c=0\) \(b=-1\) \(c=1\) Then, \(\left(a+b+c^{2}\right)=-1-1+1=-1\)
\(x^{3}-y^{2}=0\) \(3 x^{2}-2 y \frac{d y}{d x}=0\) $\frac{d y}{d x}=\frac{3 x^{2}}{2 y}=\frac{3 \times 16 m^{4}}{2 \times 8 m^{3}}=3 m$ Slope of normal \(=\frac{-1}{3 \mathrm{~m}_{1}}\) at point \(\mathrm{m}_{1}\) \(y-8 m^{3}=3 m\left(x-4 m^{2}\right) \rightarrow\) eqn of tangent \(y-8 m_{1}^{3}=\frac{-1}{3 m_{1}}\left(x-4 m_{1}^{2}\right) \rightarrow\) eqn of normal. \(\Rightarrow \frac{3 m}{-1 / 3 m_{1}}=1\) \(\Rightarrow 9 \mathrm{~mm}_{1}=-1\) Now, \(\frac{8 m^{3}-12 m^{3}}{-4 m^{3}}=8 m_{1}^{3}+\frac{4}{3} m_{1}\) \(=\frac{-8}{(9 \mathrm{~m})^{3}}-\frac{4}{27 \mathrm{~m}}\) \(\Rightarrow+4 m^{6}=\frac{8}{9^{3}}+\frac{4}{27} m^{2}\) \(=\frac{8+108 \mathrm{~m}^{2}}{9^{3}}\)
\(y^{2}-2 x^{2}-4 y+8=0\) \(2 y \frac{d y}{d x}-4 x-4 \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{2 x}{y-2}\) Now, \(y^{2}-2 x^{2}-4 y+8=0\) \((y-2)^{2}-2\left(x^{2}-2\right)=0\) eqn of tangent \(y-y_{1}=\frac{2 x_{1}}{y_{1}-2}\left(x-x_{1}\right)\) \(-\left(2-y_{1}\right)^{2}=2 x_{1}\left(1-x_{1}\right)\) Using (II) \(-2\left(x_{1}^{2}-2\right)=2 x_{1}-2 x_{1}^{2}\) \(2 \mathrm{x}_{1}=4\) \(\mathrm{x}_{1}=2\) \& \(\mathrm{x}_{1}=0\) (Horizontal tangent)
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