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Chapter 5: Problem 2

\(x+y=x^{y}\) cuts \(x\)-axis at \((1,0)\) Now, \(1+y^{\prime}=x^{y}\left(\frac{y}{x}+y^{\prime} \ln x\right)\) \(y^{\prime}=-1\) Slope of normal \(=1\) eqn of normal \(y=x-1\)

Short Answer

Expert verified
Answer: The given equation cuts the x-axis at the point (1,0), and the equation of the normal line to the curve at that point is \(y=x-1\).

Step by step solution

01

Determine the point where the equation cuts x-axis

To find the point where the equation cuts the x-axis, we need to set y equal to 0 and solve for x. So, \(x+0=x^{0} \Rightarrow x+0=1 \Rightarrow x=1\) Hence, the given equation cuts the x-axis at the point \((1,0)\).
02

Find the derivative of y with respect to x

To find the derivative of y with respect to x, we can apply implicit differentiation on the given equation. Differentiating both sides with respect to x: \((x+y)'=(x^{y})' \Rightarrow 1+y'=(yx^{y-1})'\) Applying the Product Rule and the Chain Rule, we get: \(1+y'=y'x^{y-1}+\ln x\cdot y\cdot x^{y-1}\) Now, given that \(y'=-1\) at the point \((1,0)\). Substitute the point and the value of \(y'\) into the equation: \(1-1= -1 \cdot (1)^{0-1}+\ln 1\cdot 0 \cdot (1)^{-1} \Rightarrow 0=0\) This value satisfies the equation, so it is correct.
03

Find the slope of the normal

Since the slope of the tangent is \(-1\), the slope of the normal will be the negative reciprocal of the tangent's slope. So, the slope of the normal is 1.
04

Find the equation of the normal

Since we have the slope of the normal line and the point \((1,0)\) where the normal passes through, we can find the equation of the normal using the point-slope form of a linear equation: \(y-y_{1}=m(x-x_{1})\) where \((x_{1},y_{1})=(1,0)\) and \(m=1\) (slope of the normal). So, \(y-0=1(x-1) \Rightarrow y=x-1\) Therefore, the equation of the normal line to the curve at the point \((1,0)\) is \(y=x-1\).

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Most popular questions from this chapter

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

\(y=\frac{x^{2}}{4}-2\) \(\frac{d y}{d x}=\frac{x}{2}\) \(\left(y-\frac{x_{\perp}^{2}}{4}+2\right)=\frac{-2}{x}\left(x-x_{1}\right)\) \(-1-\frac{x_{1}^{2}}{4}+2=\frac{-2}{x_{1}}\left(1-x_{1}\right)\) \(\Rightarrow 2 x_{1}-\frac{x_{1} 3}{4}=-2+2 x_{1}\) \(x_{1}^{3}=8\) \(x_{1}=2, \quad y=-1\)

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

\(y=x^{3}+a x\) \(\frac{d y}{d x}=3 x^{2}+a\) \(\frac{d y}{d x}=3+a\) \(y=b x^{2}+c\) \(\frac{d y}{d x}=2 b x=-2 b\) \(\Rightarrow 3+a=-2 b\) \(\Rightarrow 3+a+2 b=0\) As \(-1-a=0\) \(\Rightarrow a=-1\) \(b+c=0\) \(b=-1\) \(c=1\) Then, \(\left(a+b+c^{2}\right)=-1-1+1=-1\)

\(1^{2}=x^{2}+y^{2}\) \(1 \frac{d l}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t}\) \(y=x^{3 / 2}\) \(\frac{d y}{d t}=\frac{3}{2} x^{1 / 2} \frac{d x}{d t}\) Using (1) \& (2) $11 \sqrt{x^{2}+x^{3}}=x \frac{d x}{d t}+\frac{3}{2} x^{3 / 2} x^{1 / 2} \frac{d x}{d t}$ \(\frac{d x}{d t}=\frac{66}{3+\frac{27}{2}}=\frac{66 \times 2}{33}=4\)
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