/*! 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 61 \(x \sin y+y \sin x=\pi\) \(\s... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 61

\(x \sin y+y \sin x=\pi\) \(\sin y+x \cos y \frac{d y}{d x}+\frac{d y}{d x} \sin x+y \cos x=0\) \(\frac{d y}{d x}=\frac{-(\sin y+y \cos x)}{x \cos y+\sin x}=-1\) eqn of tangent \(\rightarrow y=-x+\pi\)

Short Answer

Expert verified
Answer: The equation of the tangent to the curve is \(y=-x+\pi\).

Step by step solution

01

Equation 1

\(a \sin b+b \sin a=\pi\) Now we need to find the value of (a, b) by solving Equation 1. #Step 2: Applying the first order derivative to the point (a, b)# We know that the slope of the tangent at the point (a, b) is -1. The first-order derivative represents the slope of the tangent at any point on the curve, and since the tangent touches the curve at (a, b), we can write the given first-order derivative as:
02

Equation 2

\(\frac{-(\sin b+b \cos a)}{a \cos b+\sin a}=-1\) Now, we will solve Equations 1 and 2 simultaneously to get the values of a and b. #Step 3: Solving Equations 1 and 2 simultaneously# We already have Equation 1:
03

Equation 1 (repeated)

\(a \sin b+b \sin a=\pi\) From Equation 2, we can write:
04

Equation 2 modified

\((\sin b+b \cos a)=-(a \cos b+\sin a)\) Add both Equation 1 and modified Equation 2:
05

Combined Equation

\((a \sin b+b \sin a) + (\sin b+b \cos a) = \pi\) #Step 4: Calculate point (a, b)# Upon solving Combined Equation:
06

Point (a, b) calculation

\(a \sin b+b \sin a + \sin b+b \cos a = \pi \Rightarrow a=-b+\pi\) Now we can plug a into the tangent equation to find the final equation of the tangent: #Step 5: Calculate the equation of the tangent# Given that the slope of the tangent is -1, we can write the tangent equation as:
07

Tangent Equation

\(y=-x+c\) Substitute (a, b) into the tangent equation:
08

Tangent Equation with (a, b)

\(b=-a+c\) Since \(a=-b+\pi\), we can write the tangent equation as:
09

Final Tangent Equation

\(y=-x+\pi\) Thus, the equation of the tangent to the curve represented by the given equation is \(y=-x+\pi\).

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Most popular questions from this chapter

\(y=x^{3}-x^{2}-x+2\) \(\frac{d y}{d x}=3 x^{2}-2 x-1=0\) at \(x=1\) eqn of tangent \(y=1\) solving with curve, \(x^{3}-x^{2}-x+1=0\) \((x-1)\left(x^{2}-1\right)=0\) \(\mathrm{x}=\pm \mathrm{l}\)

\(\mathrm{y}=\mathrm{x}^{2}+\mathrm{bx}-\mathrm{b}, \quad(1,1)\) lies on curve \(\frac{d y}{d x}=2 x+b=b+2\) eqn of tangent \(\Rightarrow y-1=(b+2)(x-1)\) \(x\) int \(\Rightarrow 1-\frac{1}{b+2}\) \(y\) int \(\Rightarrow 1-(b+2)\) Area of \(\Delta=\frac{1}{2} \frac{(b+1)^{2}}{(b+2)}\) \(\Rightarrow(b+1)^{2}=-4 b-8\) \(\Rightarrow b^{2}+6 b+9=0\) \((b+3)^{2}=0\) \(\Rightarrow b=-3\)

A) Area of \(\mathrm{ABCD}=2 \mathrm{Ar}(\triangle \mathrm{ABC})\) \(=2 \times \frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}\) \(=3 \times 2=6\) B) \(\mathrm{f}(\mathrm{x})=\frac{1}{\ln |\mathrm{x}|}\) is discontinous at \(x=0, \pm 1\) C) $f^{\prime}(x)=\lim _{n \rightarrow 0} \frac{f(x+n)-f(x)}{n}=\lim _{a \rightarrow 0} f(x)\left(\frac{f(n)-1)}{n}\right)$ \(=f(x) f^{\prime}(0)\) Put \(x=5\) \(f^{\prime}(5)=f(5) f^{\prime}(0)=6\) D) \(\tan \frac{3 \pi}{4}=-\frac{1}{f^{\prime}(3)}\) \(\Rightarrow f^{\prime}(3)=1\) \(A \rightarrow(R), B \rightarrow(Q), C \rightarrow(R), D \rightarrow(P)\)

\(y^{2}-2 y-8 x+17=0\) \(2 y \frac{d y}{d x}-2 \frac{d y}{d x}-8=0\) \(\frac{d y}{d x}=\frac{y}{y-1}=1\) \(\Rightarrow y=5, \quad x=4\)

\(y=\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-4-\frac{x_{2}^{2}}{4}\right)^{2}\) Min value of \(y\) is shortest distance between \(y=x-4\) \& \(y=\frac{x^{2}}{4}\) slope of normal \(=-1\) \(\frac{d y}{d x}=\frac{x}{2}\) \(-\frac{d x}{d y}=\frac{-2}{x}=-1\) \(\mathrm{x}=2\)
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