Chapter 5: Problem 43
\(y=1-a x^{2}\) \(\frac{d y}{d x}=-2 a x\) \& \(\mathrm{y}=\mathrm{x}^{2}\) \(\frac{d y}{d x}=2 x\) For orthogonal \(\Rightarrow 4 a x^{2}=1\) Now, \(1-a x^{2}=x^{2}\) \(x^{2}=\frac{1}{1+a}\) Using eq (1) \(\frac{4 a}{1+a}=1\) \(a=\frac{1}{3}\)
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slope of normal \(\Rightarrow 3 x-y+3=0\) $$ x=0 \& y=3 $$ Pt of normal \(=(0,3)\) \(\frac{d y}{d x}=\frac{-1}{3}=f^{\prime}(0)\) $\lim _{x \rightarrow 0} \frac{x^{2}}{f\left(x^{2}\right)+4 f\left(7 x^{2}\right)-5 f\left(4 x^{2}\right)}$ $\lim _{x \rightarrow 0} \frac{2 x}{x\left[2 f^{\prime}\left(x^{2}\right)+56 f^{\prime}\left(7 x^{2}\right)-40 f^{\prime}\left(4 x^{2}\right)\right]}$ \(=\frac{2}{-6}=\frac{-1}{3}\)
At exercise, respiratory cycle has a period of 4 . Hence difference in frequency is \(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
$y=\sin ^{-1} 2 x \sqrt{1-x^{2}}= \begin{cases}2 \sin ^{-1} x & |x| \leq \frac{1}{\sqrt{2}} \\ \pi-2 \sin ^{-1} x & x>\frac{1}{\sqrt{2}} \\\ -\left(\pi+2 \sin ^{-1} x\right) & x<\frac{-1}{\sqrt{2}}\end{cases}$ \(\mathrm{x}=0, \mathrm{y}=0\) is only integral point where the function has a unique tangent
\(\frac{d h_{4}}{d t}=0.5 t+2\) Integrating \(=\frac{t^{2}}{4}+2 t+5\) \(\frac{d h_{b}}{d t}=t+1\) Integrating, \(h_{b}=\frac{t^{2}}{2}+t+5\)
\((x-4)^{2}+y^{2}=c^{2}\) \(2(x-4)+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{4-x}{y}\) For \(y^{2}=x^{3}+1\) \(2 y \frac{d y}{d x}=3 x^{2}\) \(\frac{d y}{d x}=\frac{3 x^{2}}{2 y}\) \(\Rightarrow \frac{4-x}{y}=\frac{3 x^{2}}{2 y}\) \(\Rightarrow 3 x^{2}+2 x-8=0\) \(\Rightarrow(3 x-4)(x+2)=0\) \(x=\frac{4}{3}\) or \(-2\)
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