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Chapter 5: Problem 34
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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\(y=2 x^{2}-x, \quad y=2-x^{3}\) Pt of intersection are \(2 \mathrm{x}^{2}-\mathrm{x}=2-\mathrm{x}^{3}\) \(\Rightarrow 2\left(x^{2}-1\right)+x\left(x^{2}-1\right)=0\) \(\Rightarrow\left(x^{2}-1\right)(x+2)=0\) \(\Rightarrow x=\pm 1,-2\) \(\Rightarrow(1,1),(-1,3),(-2,10)\) For \(\mathrm{y}=2 \mathrm{x}^{2}-\mathrm{x}\) \(\frac{d y}{d x}=4 x-1\) For \(y=2-x^{3}\) \(\frac{d y}{d x}=-3 x^{2}\) At \((1,1)\) $\theta=\tan ^{-1}\left(\frac{3+3}{1-9}\right)=\tan ^{-1}\left(\frac{3}{-4}\right)$ At \((-1,3)\) $\theta=\tan ^{-1}\left(\frac{-5+3}{1+15}\right)=\tan ^{-1}\left(\frac{1}{8}\right)$ At \((-2 ; 10)\) \(\theta=\tan ^{-1}\left(\frac{-9+12}{1+108}\right)=\tan ^{-1} \frac{3}{109}\)
\(\frac{d h_{4}}{d t}=0.5 t+2\) Integrating \(=\frac{t^{2}}{4}+2 t+5\) \(\frac{d h_{b}}{d t}=t+1\) Integrating, \(h_{b}=\frac{t^{2}}{2}+t+5\)
\(y=6-x-x^{2}\) \(\frac{d y}{d x}=-1-2 x\) eqn of tangent $y-6+x_{1}+x_{1}^{2}=-\left(1+2 x_{1}\right)\left(x-x_{1}\right)$ \(x y=x+3\) \(x \frac{d y}{d x}+y=1\) \(\frac{d y}{d x}=\frac{1-y_{2}}{x}\) eqn of tangent \(\rightarrow\) \(y-\frac{x_{2}+3}{x_{2}}=\frac{1-\frac{x_{2}+3}{x_{2}}}{x_{2}}\left(x-x_{2}\right)\) \(x_{2} y-\left(x_{2}+3\right)=\frac{-3}{x_{2}}\left(x-x_{2}\right)\) Comparing the two eqns $\frac{1}{x_{2}}=\frac{-\left(1+2 x_{1}\right)}{-\frac{3}{x_{2}}}=\frac{x_{1}\left(1+2 x_{1}\right)+6-x_{1}-x_{1}^{2}}{3+x_{2}+3}$ $\frac{1}{x_{2}}=\frac{x_{2}\left(1+2 x_{1}\right)}{3} \& \frac{1}{x_{2}}=\frac{x_{1}^{2}+6}{x_{2}+6}$ \(3=x_{2}^{2}+2 x_{1} x_{2}^{2} \quad\) \& \(x_{2}+6=x_{1}^{2} x_{2}+6 x_{2}\) $3=\left(1+2 \mathrm{x}_{1}\right) \mathrm{x}_{2}^{2} \quad \& \quad \mathrm{x}_{2}=\frac{6}{5-\mathrm{x}_{1}^{2}}$ \(\Rightarrow 3=\frac{\left(1+2 x_{1}\right) 36}{\left(5-x_{1}^{2}\right)^{2}}\) $\Rightarrow\left(5-\mathrm{x}_{1}^{2}\right)^{2}=12\left(1+2 \mathrm{x}_{1}\right)$ Let \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) be the common tangent \(m x+c=6-x-x^{2}\) \(\Rightarrow \mathrm{x}^{2}+(\mathrm{m}+1) \mathrm{x}+(\mathrm{c}-6)=0\) \(\mathrm{D}=0\) \((m+1)^{2}-4(c-6)=0\) For \(x y=x+3\) \(x(m x+c)=x+3\) \(m x^{2}+(c-1) x-3=0\) \(\mathrm{D}=0\) \((c-1)^{2}+12 m=0\) Using (I) \& (II) \((m+1)^{2}+24=4 c\) \(\mathrm{c}-1=\frac{(\mathrm{m}+1)^{2}+20}{4}\) $\Rightarrow\left(\frac{(\mathrm{m}+\mathrm{l})^{2}+20}{4}\right)^{2}+12 \mathrm{~m}=0$ \(\left((m+1)^{2}+20\right)^{2}+192 m=0\) Upon solving, \(\mathrm{m}=-3\) \(\Rightarrow \mathrm{c}=7\) eqn of tangent \(=y=-3 x+7\)
\(\mathrm{l}_{1}=\mathrm{l}_{2}+\mathrm{l}_{2}^{3}+6\) $\frac{\mathrm{d} \mathrm{l}_{1}}{\mathrm{dt}}=\frac{\mathrm{dl}_{2}}{\mathrm{dt}}+3 \mathrm{l}_{2}^{2} \frac{\mathrm{dl}_{2}}{\mathrm{dt}}$ $\frac{\mathrm{d} \mathrm{S}_{2}}{\mathrm{dS}_{1}}=\frac{\mathrm{dS}_{2}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dS}_{1}}$ $=\frac{2 \mathrm{l}_{2} \mathrm{~d}_{2} / \mathrm{dt}}{2 \mathrm{l}_{1} \mathrm{dl}_{1} / \mathrm{dt}}$ \(=\frac{1}{4} \times \frac{1}{8}\)
\(y=e^{2 x}+x^{2}\) \(\frac{d y}{d x}=2 e^{2 x}+2 x\) at \(x=0, \quad \frac{d y}{d x}=2\) \(-\frac{d x}{d y}=-\frac{1}{2}\) eqn of normal \(\rightarrow y-1=-\frac{1}{2} x\) \(\Rightarrow 2 y+x-2=0\) Distance from \((0,0)=\frac{2}{\sqrt{5}}\)
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