chapter review exercises\n1. the position of a particle at time $t(s)$ is $s(t)=sqrt{t^{2}+1}$ m. compute…

Question 1

Explanation:

Step1: Recall average - velocity formula

The average velocity of a particle with position function $s(t)$ over the interval $[a,b]$ is $v_{avg}=\frac{s(b)-s(a)}{b - a}$. Given $s(t)=\sqrt{t^{2}+1}$, $a = 2$, and $b = 5$. First, find $s(2)$ and $s(5)$: $s(2)=\sqrt{2^{2}+1}=\sqrt{4 + 1}=\sqrt{5}$ $s(5)=\sqrt{5^{2}+1}=\sqrt{25+1}=\sqrt{26}$ Then $v_{avg}=\frac{s(5)-s(2)}{5 - 2}=\frac{\sqrt{26}-\sqrt{5}}{3}\approx\frac{5.099 - 2.236}{3}=\frac{2.863}{3}\approx0.95$ m/s.

Step2: Recall instantaneous - velocity formula

The instantaneous velocity $v(t)$ is the derivative of the position function $s(t)$. Using the chain - rule, if $s(t)=(t^{2}+1)^{\frac{1}{2}}$, then $s^\prime(t)=\frac{1}{2}(t^{2}+1)^{-\frac{1}{2}}\cdot2t=\frac{t}{\sqrt{t^{2}+1}}$. To find the instantaneous velocity at $t = 2$, substitute $t = 2$ into $s^\prime(t)$: $s^\prime(2)=\frac{2}{\sqrt{2^{2}+1}}=\frac{2}{\sqrt{5}}\approx\frac{2}{2.236}\approx0.89$ m/s.

Question 2

Explanation:

Step1: Recall instantaneous - velocity formula

The instantaneous velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=15.2t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, where $a = 15.2$ and $n = 2$. So $v(t)=s^\prime(t)=30.4t$.

Step2: Evaluate at $t = 5$

Substitute $t = 5$ into $v(t)$: $v(5)=30.4\times5 = 152$ m/s.

Question 3

Explanation:

Step1: Recall slope of secant - line formula

The slope of the secant line between two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve $y = f(x)$ is $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Given $f(x)=\sqrt{2x}$, $x_1 = 16$. For $x_2=16 + 0.01$: $f(16)=\sqrt{2\times16}=\sqrt{32}=4\sqrt{2}$ $f(16 + 0.01)=\sqrt{2\times(16 + 0.01)}=\sqrt{32.02}$ $m_1=\frac{\sqrt{32.02}-4\sqrt{2}}{(16 + 0.01)-16}=\frac{\sqrt{32.02}-4\sqrt{2}}{0.01}\approx\frac{5.6586 - 5.6569}{0.01}=\frac{0.0017}{0.01}=0.17$ Similarly, for $x_2=16-0.01$: $m_2=\frac{\sqrt{31.98}-4\sqrt{2}}{(16 - 0.01)-16}=\frac{\sqrt{31.98}-4\sqrt{2}}{-0.01}\approx0.17$ For $x_2=16 + 0.001$: $m_3=\frac{\sqrt{32.002}-4\sqrt{2}}{0.001}\approx0.177$ For $x_2=16-0.001$: $m_4=\frac{\sqrt{31.998}-4\sqrt{2}}{-0.001}\approx0.177$ For $x_2=16 + 0.0001$: $m_5=\frac{\sqrt{32.0002}-4\sqrt{2}}{0.0001}\approx0.177$ For $x_2=16-0.0001$: $m_6=\frac{\sqrt{31.9998}-4\sqrt{2}}{-0.0001}\approx0.177$ The slope of the tangent line at $x = 16$ is approximately $0.18$.

Question 4

Explanation:

Step1: Recall slope of secant - line formula

The slope of the secant line between two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve $y = f(x)$ is $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Given $f(x)=x^{3}-2x$, $x_1 = 5$, and $x_2=x$. $f(5)=5^{3}-2\times5=125 - 10 = 115$ $f(x)=x^{3}-2x$ $m=\frac{(x^{3}-2x)-115}{x - 5}=\frac{x^{3}-2x-115}{x - 5}$ By polynomial long - division or synthetic division, $\frac{x^{3}-2x-115}{x - 5}=x^{2}+5x + 23$.

Step2: Recall slope of tangent - line formula

The slope of the tangent line at $x = 5$ is the limit of the slope of the secant line as $x$ approaches $5$. Substitute $x = 5$ into $x^{2}+5x + 23$: $m=5^{2}+5\times5+23=25 + 25+23=73$.

Question 5

Explanation:

Step1: Use L'Hopital's rule (since it is in $\frac{0}{0}$ form)

As $x\rightarrow0$, $\frac{1-\cos^{3}(x)}{x^{2}}$ is in the $\frac{0}{0}$ form (because $\cos(0)=1$, so $1-\cos^{3}(0)=0$ and $x^{2}=0$ when $x = 0$). Differentiate the numerator and denominator. The derivative of the numerator $u = 1-\cos^{3}(x)$ using the chain - rule: $u^\prime=3\cos^{2}(x)\sin(x)$. The derivative of the denominator $v=x^{2}$ is $v^\prime = 2x$. So $\lim_{x\rightarrow0}\frac{1-\cos^{3}(x)}{x^{2}}=\lim_{x\rightarrow0}\frac{3\cos^{2}(x)\sin(x)}{2x}$. This is still in the $\frac{0}{0}$ form. Apply L'Hopital's rule again. The derivative of the numerator $3\cos^{2}(x)\sin(x)$ using the product - rule: $(3\cos^{2}(x)\sin(x))^\prime=3(-2\cos(x)\sin^{2}(x)+\cos^{3}(x))$. The derivative of the denominator $2x$ is $2$. $\lim_{x\rightarrow0}\frac{3(-2\cos(x)\sin^{2}(x)+\cos^{3}(x))}{2}=\frac{3(0 + 1)}{2}=1.50$.

Question 6

Explanation:

Step1: Let $y=x^{\frac{1}{x - 1}}$, then $\ln y=\frac{\ln x}{x - 1}$

As $x\rightarrow1$, $\frac{\ln x}{x - 1}$ is in the $\frac{0}{0}$ form (since $\ln(1)=0$ and $x - 1=0$ when $x = 1$). Apply L'Hopital's rule. The derivative of the numerator $\ln x$ is $\frac{1}{x}$, and the derivative of the denominator $x - 1$ is $1$. So $\lim_{x\rightarrow1}\frac{\ln x}{x - 1}=\lim_{x\rightarrow1}\frac{1}{x}=1$. Since $\lim_{x\rightarrow1}\ln y = 1$, then $\lim_{x\rightarrow1}y=e^{1}\approx2.72$.

Question 7

Explanation:

Step1: Factor the denominator

As $x\rightarrow2$, $\frac{x^{x}-4}{x^{2}-4}$ is in the $\frac{0}{0}$ form (since $2^{2}=4$ and $x^{2}-4=(x - 2)(x + 2)$). We know that $x^{x}=e^{x\ln x}$. Using L'Hopital's rule, the derivative of the numerator $u = x^{x}=e^{x\ln x}$, and by the chain - rule and product - rule, $u^\prime=x^{x}(\ln x + 1)$. The derivative of the denominator $v=x^{2}-4$ is $v^\prime = 2x$. $\lim_{x\rightarrow2}\frac{x^{x}-4}{x^{2}-4}=\lim_{x\rightarrow2}\frac{x^{x}(\ln x + 1)}{2x}=\frac{2^{2}(\ln 2+1)}{2\times2}=\frac{4(\ln 2 + 1)}{4}=\ln 2+1\approx0.69+1 = 1.69$.

Question 8

Explanation:

Step1: Use L'Hopital's rule (since it is in $\frac{0}{0}$ form)

As $x\rightarrow2$, $\frac{x - 2}{\ln(3x - 5)}$ is in the $\frac{0}{0}$ form (because when $x = 2$, $x - 2=0$ and $\ln(3\times2-5)=\ln(1)=0$). Differentiate the numerator and denominator. The derivative of the numerator $u=x - 2$ is $u^\prime = 1$. The derivative of the denominator $v=\ln(3x - 5)$ using the chain - rule is $v^\prime=\frac{3}{3x - 5}$. $\lim_{x\rightarrow2}\frac{x - 2}{\ln(3x - 5)}=\lim_{x\rightarrow2}\frac{1}{\frac{3}{3x - 5}}=\lim_{x\rightarrow2}\frac{3x - 5}{3}=\frac{3\times2-5}{3}=\frac{1}{3}\approx0.33$.

Question 9

Explanation:

Step1: First, find a common denominator

$\frac{7}{x - 7}-\frac{3}{x - 3}=\frac{7(x - 3)-3(x - 7)}{(x - 7)(x - 3)}=\frac{7x-21-3x + 21}{(x - 7)(x - 3)}=\frac{4x}{(x - 7)(x - 3)}$. As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{4x}{(x - 7)(x - 3)}=\lim_{x\rightarrow\infty}\frac{4x}{x^{2}-10x + 21}$. Divide both the numerator and denominator by $x^{2}$: $\lim_{x\rightarrow\infty}\frac{\frac{4}{x}}{1-\frac{10}{x}+\frac{21}{x^{2}}}=0$.

Answer:

1. Average velocity: $\approx0.95$ m/s, Instantaneous velocity at $t = 2$: $\approx0.89$ m/s
2. Instantaneous velocity at $t = 5$: $152$ m/s
3. Slope of tangent line at $x = 16$: $\approx0.18$
4. Slope of tangent line at $x = 5$: $73$
5. $\lim_{x\rightarrow0}\frac{1-\cos^{3}(x)}{x^{2}}=1.50$
6. $\lim_{x\rightarrow1}x^{\frac{1}{x - 1}}\approx2.72$
7. $\lim_{x\rightarrow2}\frac{x^{x}-4}{x^{2}-4}\approx1.69$
8. $\lim_{x\rightarrow2}\frac{x - 2}{\ln(3x - 5)}\approx0.33$
9. $\lim_{x\rightarrow\infty}(\frac{7}{x - 7}-\frac{3}{x - 3})=0$