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Chapter 5: Problem 12
\(P(t)=60 t^{2}-t^{3}\) \(P^{\prime}(t)=120 t-3 t^{2}=900\) \(\Rightarrow t^{2}-40 t+300=0\) \(t=10,30\)
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\(f^{\prime}(x)= \begin{cases}-2 x, & x<0 \\ 2 x, & x \geq 0\end{cases}\) eqn of tangent $\rightarrow \quad y-y_{1}=2 x_{1}\left(x-x_{1}\right), x \geq 0$ \(y-y_{2}=-2 x_{2}\left(x-x_{2}\right) x<0\)
\(y=-t+e^{a t}=0\) \(x=t+e^{a}\) \(\frac{d y}{d t}=-1+a e^{u r}\) \(\frac{d x}{d t}=1+a e^{u}\) \(\frac{d y}{d x}=\frac{a e^{a t}-1}{a e^{a t}+1}=0\) \(\Rightarrow \mathrm{e}^{a t}=\frac{1}{a}\) \(\Rightarrow a t=-\ln a\) \(\mathrm{t}=-\frac{\ln \mathrm{a}}{\mathrm{a}}\) $\mathrm{x}=-\frac{\ln \mathrm{a}}{\mathrm{a}}+\frac{1}{\mathrm{a}}=\frac{\ln \mathrm{e} / \mathrm{a}}{\mathrm{a}}=\frac{2}{\mathrm{a}}$
\(x^{2}+y^{2}-\frac{10}{3} y+1=0\) \(x^{2}+\left(y-\frac{5}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}\) \(y^{2}=x^{3}\) \(2 y y_{1}=3 x^{2}\) \(y_{1}=\frac{3 x^{2}}{2 y}\) Normal eqn \(\quad y-y_{1}=-\frac{2 y_{1}}{3 x_{1}^{2}}\left(x-x_{1}\right)\) as it passes through \(\left(0, \frac{5}{3}\right)\) \(\frac{5}{3}-y_{1}=\frac{2 y_{1}}{3 x_{1}}\) \(5-3 y_{1}=2 y^{1 / 3}\) \(5-9 y^{3}-225 y+135 y^{2}=8 y\) \(9 y^{3}+233 y-135 y^{2}-25=0\)
\(y=1-a x^{2}\) \(\frac{d y}{d x}=-2 a x_{1}\) \& \(y=x^{2}\) \(\frac{d y}{d x}=2 x_{1}\) If curves are orthogonal, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(+4 a x^{2}=1\) Now \(1-a x^{2}=x^{2} \Rightarrow \frac{1}{1+a}=x^{2}\) Using eq (1), we get \(\frac{4 a}{1+a}=1\) \(\mathrm{a}=\frac{1}{3}\)
Slope of secant \(=\frac{9 a^{2}-a^{2}}{3 a-a}=4 a\) \(\frac{d y}{d x}=2 x=4 a\) \(x=2 a\) \(y=4 a^{2}\)
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