/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 \(y=\ln x\) \(\frac{d y}{d x}=... [FREE SOLUTION] | 91影视

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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

\(y^{2}-2 x^{2}-4 y+8=0\) \(2 y \frac{d y}{d x}-4 x-4 \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{2 x}{y-2}\) 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)

\(y=\left(\frac{x}{2}-a\right)^{2}+a-2\) \(4(y-(a-2))=(x-2 a)^{2}\) Vertex \(\Rightarrow \mathrm{h}=2 \mathrm{a}, \mathrm{k}=\mathrm{a}-2\) Locus of vertex \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}-2\) \(\Rightarrow 2 y=x-4\)

\(x^{2}+2 x y+2 y^{2}=45\) \(x+y+x \frac{d y}{d x}+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-(x+y)}{x+2 y}=-2\) \(\Rightarrow x+y=2 x+4 y\) \(\Rightarrow x+3 y=0\) Put eq (II) in eq (I) \(9 y^{2}-6 y^{2}+2 y^{2}=45\) \(y=\pm 3\) \(x=\pm 9\) Pt \(\rightarrow(9,-3), \quad(-9,3)\)

\(\alpha y^{2}=(x+\beta)^{3}\) \(2 \alpha y \frac{d y}{d x}=3(x+\beta)^{2}\) \(y \frac{d y}{d x}=\frac{3(x+\beta)^{2}}{2 \alpha}=\) Sub Normal \(y \frac{d x}{d y}=\frac{2 \alpha y^{2}}{3(x+\beta)^{2}}=\) SubTangent Square of sub tangent \(=\frac{4 \alpha^{2}(x+\beta)^{4}}{9}\)

\(|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}\) \(|\mathbf{x}|-\mathrm{a}>1 \quad\) for no solution \(|x|>1+a\)
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