/*! 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 69 \(\frac{d h_{4}}{d t}=0.5 t+2\) ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 69

\(\frac{d h_{4}}{d t}=0.5 t+2\) Integrating \(=\frac{t^{2}}{4}+2 t+5\) \(\frac{d h_{b}}{d t}=t+1\) Integrating, \(h_{b}=\frac{t^{2}}{2}+t+5\)

Short Answer

Expert verified
Answer: The original functions are \(h_{4}(t) = \frac{t^2}{4}+2t+C_{1}\) and \(h_{b}(t) = \frac{t^2}{2}+t+C_{2}\), where \(C_{1}\) and \(C_{2}\) are constants of integration.

Step by step solution

01

Integrate \(\frac{dh_{4}}{dt}\) with respect to \(t\)

We are given \(\frac{dh_{4}}{dt}=0.5t+2\) and its integral, \(h_{4}(t)=\frac{t^2}{4}+2t+C_{1}\). Note that constant of integration, \(C_{1}\), has been added to the integral. #Step 2: Integrate the first derivative of \(h_{b}(t)\)#
02

Integrate \(\frac{dh_{b}}{dt}\) with respect to \(t\)

We are given \(\frac{dh_{b}}{dt}=t+1\). To find \(h_{b}(t)\), we need to integrate this expression with respect to \(t\). We obtain \(h_{b}(t)=\int(t+1)dt = \frac{t^{2}}{2}+t+C_{2}\), where \(C_{2}\) is the constant of integration. #Step 3: Write down the functions \(h_{4}(t)\) and \(h_{b}(t)\)#
03

State the final expressions for \(h_{4}(t)\) and \(h_{b}(t)\)

Utilizing the integrals obtained in Steps 1 and 2, the original functions are: - \(h_{4}(t) = \frac{t^2}{4}+2t+C_{1}\) - \(h_{b}(t) = \frac{t^2}{2}+t+C_{2}\) However, since the constant terms \(C_{1}\) and \(C_{2}\) cannot be determined without further information, this is the most simplified form of the original functions that we can provide.

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Most popular questions from this chapter

\(x^{4}+y^{4}=a^{4}\) \(x^{3}+y^{3} \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x^{3}}{y^{3}}\) eqn of tangent \(\rightarrow\) \(\mathrm{y}-\mathrm{y}_{1}=-\frac{\mathrm{x}_{1}^{3}}{\mathrm{y}_{1}^{3}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{p}=\frac{\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}+\mathrm{x}_{1}=\frac{\mathrm{x}_{1}^{4}+\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{x}_{1}^{3}}\) \(\mathrm{q}=\mathrm{y}_{1}+\frac{\mathrm{x}_{1}^{4}}{\mathrm{y}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{y}_{1}^{3}}\) Now, \(\mathrm{p}^{-4 / 3}+\mathrm{q}^{-4 / 3}=\mathrm{a}^{-4 / 3}\)

\(f(0)<0\) \(\Rightarrow \mathrm{a}^{2}+\mathrm{a}-2<0\) \(\Rightarrow a \in(-2,1)\) Integral values of a are \(-1,0\).

Let \(y=m x+\frac{1}{m}\) be tangent to \(y^{2}=4 x\) eqn of normal at \(\left(x_{1}, y_{1}\right)\) to \(x^{2}=4\) by is \(y-y_{1}=-\frac{2 b}{x_{1}}\left(x-x_{1}\right)\) \(\Rightarrow y=-\frac{2 b}{x_{1}} x+\frac{x_{1}^{2}}{4 b}+2 b\) Comparing two eqn \(\mathrm{m}=-\frac{2 \mathrm{~b}}{\mathrm{x}_{1}}\) \(\frac{\mathrm{x}_{1}^{2}}{4 \mathrm{~b}}+2 \mathrm{~b}=\frac{1}{\mathrm{~m}}\) Using (I) \& (II)

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

\(y=1-a x^{2}\) \(\frac{d y}{d x}=-2 a x\) \& \(\mathrm{y}=\mathrm{x}^{2}\) \(\frac{d y}{d x}=2 x\) For orthogonal \(\Rightarrow 4 a x^{2}=1\) Now, \(1-a x^{2}=x^{2}\) \(x^{2}=\frac{1}{1+a}\) Using eq (1) \(\frac{4 a}{1+a}=1\) \(a=\frac{1}{3}\)
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