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Chapter 5: Problem 21

\(y=\ln x\) \(\frac{d y}{d x}=\frac{1}{x}\) Slope of normal \(\Rightarrow-x_{1}\) Slope of 1 chord \(=\frac{1}{e-1}\) \(\Rightarrow \frac{x_{1}}{e-1}=1\) \(\Rightarrow \quad x_{1}=e-1\) \(y_{1}=\ln (e-1)\)

Short Answer

Expert verified
In summary, the point at which the normal intersects the curve is given by the coordinates \((x_1, y_1) = (e - 1, \ln (e - 1))\).

Step by step solution

01

Find the derivative of y with respect to x

To find the derivative \(\frac{dy}{dx}\), we'll use the differentiation rules for logarithmic functions: \(\frac{d}{dx}\ln{x}=\frac{1}{x}\). So, in this case, we have \(\frac{dy}{dx}=\frac{1}{x}\).
02

Find the slope of the normal

Since we are given that the slope of the normal is \(-x_1\), we can directly use this expression for further calculations.
03

Find the slope of the 1-chord

We are given that the slope of the 1-chord is \(\frac{1}{e-1}\). This value will be used to determine \(x_1\).
04

Equate the slopes and solve for \(x_1\)

Since the normal is perpendicular to the 1-chord, the product of their slopes equals -1. Therefore, we have: $$-x_1 \cdot \frac{1}{e - 1} = -1$$ Solve for \(x_1\): $$x_1 = e - 1$$
05

Find the corresponding \(y_1\)

We have found the \(x_1\) value, now we need to find the corresponding \(y_1\) value using the given function \(y = \ln x\): $$y_1 = \ln (x_1) = \ln (e - 1)$$
06

Final answer

We have found the coordinates of the point \((x_1, y_1)\) where the normal intersects the curve: $$x_1 = e - 1$$ $$y_1 = \ln (e - 1)$$

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Most popular questions from this chapter

\(y=2 x^{2}-x, \quad y=2-x^{3}\) 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 \((-2 ; 10)\) \(\theta=\tan ^{-1}\left(\frac{-9+12}{1+108}\right)=\tan ^{-1} \frac{3}{109}\)

\(y=x \ln x+\sin (\pi \ln x)\) \(\frac{d y}{d x}=1+\ln x+\cos (\pi \ln x) \frac{\pi}{x}\) at \(x=e\) \(\frac{d y}{d x}=2-\frac{\pi}{e}=\frac{2 e-\pi}{e}\) at \(x=e^{2}\) \(\frac{d y}{d x}=3+\frac{\pi}{e^{2}}=\frac{3 e^{2}+\pi}{e^{2}}\)

Volume inhaled while exercising \(=\int_{0}^{4} 1.75 \sin \frac{\pi t}{2}\) dt \(=-1.75\left(\cos \frac{\pi t}{2}\right) \times \frac{2}{\pi}\) \(=\frac{7}{\pi}\) Difference \(=\frac{7}{\pi}-\frac{5.1}{\pi}\) \(=\frac{1.9}{\pi}\) Hence, B is correct Comprehension \(5:\) $\begin{aligned} x=& a(2 \cos t+\cos 2 t), \quad y=a(2 \sin t-\sin 2 t\\\ \frac{d x}{d t} &=a(-2 \sin t-2 \sin 2 t) \\\ &=-2 a[\sin t+2 \sin 2 t] \\ &=-2 a \sin t[1+2 \cos t] \\ \frac{d y}{d t} &=a[2 \cos t-2 \cos 2 t] \\ &=-2 a\left[2 \cos ^{2} t-\cos t-1\right] \\\ &=-2 a(\cos t-1)(2 \cos t+1) \end{aligned}$ $\frac{d y}{d x}=\frac{(\cos t-1)(2 \cos t+1)}{(\sin t)(1+2 \cos t)}=-\tan t / 2$ slope of normal \(=\cot t / 2 .\)

\(y=6-x-x^{2}\) \(\frac{d y}{d x}=-1-2 x\) 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 \(=y=-3 x+7\)

\(y=2 x^{2}-x+1\) \(\frac{d y}{d x}=4 x_{1}-1\) \(4 x_{1}-1=3\) \(x_{1}=1\) \(y_{1}=2\)
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