/*! 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 38 \(y^{3}=27 x\) \(3 y^{2} \frac... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 38

\(y^{3}=27 x\) \(3 y^{2} \frac{d y}{d x}=27 \frac{d x}{d t}\) \(\frac{3 y^{2}}{27}<1\) \(y^{2}<9 \Rightarrow y \in(-3,3)\)

Short Answer

Expert verified
Answer: The relationship between \(dy/dx\) and \(dy/dt\) is \(\frac{dy}{dx} = \frac{dy}{dt}\) when \(y \in (-3, 3)\).

Step by step solution

01

Express x in terms of y

Since \(y^3 = 27x\), we can write \(x\) in terms of \(y\) by dividing both sides by \(27\): \begin{align*} x= \frac{y^{3}}{27}. \end{align*}
02

Differentiate x with respect to t

Now we need to differentiate \(x\) with respect to \(t\). Using the chain rule, we have: \begin{align*} \frac{dx}{dt} = \frac{d}{dt} \left( \frac{y^{3}}{27} \right) = \frac{3y^2}{27} \frac{dy}{dt}. \end{align*}
03

Write the second equation in terms of \(dy/dx\) and \(dx/dt\)

We are given the equation: \begin{align*} 3y^2\frac{dy}{dx}=27\frac{dx}{dt}. \end{align*} By dividing both sides by \(3y^2\), we can solve for \(\frac{dy}{dx}\): \begin{align*} \frac{dy}{dx} = \frac{27\frac{dx}{dt}}{3y^2}. \end{align*}
04

Solve for \(\frac{dy}{dx}\) using the expressions from Steps 2 and 3

From Step 2, we have: \begin{align*} \frac{dx}{dt} = \frac{3y^2}{27} \frac{dy}{dt}. \end{align*} Substitute this into the equation from Step 3: \begin{align*} \frac{dy}{dx} = \frac{27\left(\frac{3y^2}{27}\frac{dy}{dt}\right)}{3y^2}. \end{align*} Simplify the equation to obtain \(\frac{dy}{dx}\): \begin{align*} \frac{dy}{dx} = \frac{dy}{dt}. \end{align*}
05

Conclude the result

We found that \(\frac{dy}{dx} = \frac{dy}{dt}\), and since \(y^2 < 9 \Rightarrow y \in (-3, 3)\), the slope of \(y\) with respect to \(x\) is the same as the slope of \(y\) with respect to \(t\) when \(y\) is within that interval.

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Most popular questions from this chapter

\(y^{2}=x\left(2-x^{2}\right)\) \(2 y \frac{d y}{d x}=2-3 x^{2}\) \(\frac{d y}{d x}=\frac{-1}{2} \quad\) at \((1,1)\) eqn of tangent \(y-1=\frac{-1}{2}(x-1)\) \(2 y+x-3=0\) Solving with curve \(\left(\frac{3-x}{2}\right)^{2}=2 x-x^{3}\) \(4 x^{3}+x^{2}-14 x+9=0\)

\(\mathrm{y}=\mathrm{x}^{2}+\mathrm{bx}-\mathrm{b}, \quad(1,1)\) 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 \(\Delta=\frac{1}{2} \frac{(b+1)^{2}}{(b+2)}\) \(\Rightarrow(b+1)^{2}=-4 b-8\) \(\Rightarrow b^{2}+6 b+9=0\) \((b+3)^{2}=0\) \(\Rightarrow b=-3\)

\(f(x)=\frac{x}{1-x^{2}}\) $f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$ \(1+x^{2}=1+x^{4}-2 x^{2}\) \(\Rightarrow \quad x^{2}\left(x^{2}-3\right)=0\) \(\Rightarrow x=0, \pm \sqrt{3}\) $P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right)\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

slope of normal \(\Rightarrow 3 x-y+3=0\) $$ x=0 \& y=3 $$ Pt of normal \(=(0,3)\) \(\frac{d y}{d x}=\frac{-1}{3}=f^{\prime}(0)\) $\lim _{x \rightarrow 0} \frac{x^{2}}{f\left(x^{2}\right)+4 f\left(7 x^{2}\right)-5 f\left(4 x^{2}\right)}$ $\lim _{x \rightarrow 0} \frac{2 x}{x\left[2 f^{\prime}\left(x^{2}\right)+56 f^{\prime}\left(7 x^{2}\right)-40 f^{\prime}\left(4 x^{2}\right)\right]}$ \(=\frac{2}{-6}=\frac{-1}{3}\)

\(y=e^{2 x}+x^{2}\) \(\frac{d y}{d x}=2 e^{2 x}+2 x\) 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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