a particle moves along a horizontal line. its position function is s(t) for t≥0. for each problem, find the…

Explanation:

Step1: Recall the relationship between position, velocity and direction - change

The velocity function $v(t)$ is the derivative of the position function $s(t)$. The particle changes directions when $v(t)=0$ and $v(t)$ changes sign. For $s(t)=t^{3}-23t^{2}+120t$, find $v(t)$ using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$. $v(t)=s^\prime(t)=3t^{2}-46t + 120$

Step2: Solve the quadratic equation for $v(t) = 0$

Set $3t^{2}-46t + 120=0$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 3$, $b=-46$, $c = 120$. $t=\frac{46\pm\sqrt{(-46)^{2}-4\times3\times120}}{2\times3}=\frac{46\pm\sqrt{2116 - 1440}}{6}=\frac{46\pm\sqrt{676}}{6}=\frac{46\pm26}{6}$ $t_1=\frac{46 + 26}{6}=\frac{72}{6}=12$ and $t_2=\frac{46-26}{6}=\frac{20}{6}=\frac{10}{3}$

Step3: Analyze the sign of $v(t)$ for the intervals

We have three intervals to consider: $0\leq t<\frac{10}{3}$, $\frac{10}{3}<t<12$, and $t > 12$. Choose test points: for $0\leq t<\frac{10}{3}$, let $t = 1$. Then $v(1)=3\times1^{2}-46\times1 + 120=3-46 + 120 = 77>0$. For $\frac{10}{3}<t<12$, let $t = 5$. Then $v(5)=3\times5^{2}-46\times5 + 120=75-230 + 120=-35<0$. For $t>12$, let $t = 13$. Then $v(13)=3\times13^{2}-46\times13 + 120=507-598+120=29>0$. The particle changes directions at $t=\frac{10}{3}$ and $t = 12$.

For $s(t)=t^{3}-13t^{2}$, find $v(t)$ and $a(t)$ (acceleration). $v(t)=s^\prime(t)=3t^{2}-26t=t(3t - 26)$ $a(t)=v^\prime(t)=6t-26$

Step4: Find when $v(t)=0$

Set $v(t)=0$, then $t(3t - 26)=0$. So $t = 0$ or $t=\frac{26}{3}$

Step5: Analyze the signs of $v(t)$ and $a(t)$

Intervals are $0\leq t<\frac{26}{3}$ and $t>\frac{26}{3}$. For $v(t)$: when $0\leq t<\frac{26}{3}$, if $t = 1$, $v(1)=3\times1^{2}-26\times1=-23<0$. When $t>\frac{26}{3}$, if $t = 9$, $v(9)=3\times9^{2}-26\times9=243-234 = 9>0$. For $a(t)$: $a(t)=6t-26$. When $t<\frac{13}{3}$, $a(t)<0$. When $t>\frac{13}{3}$, $a(t)>0$. The particle is speeding up when $v(t)$ and $a(t)$ have the same sign.

• Speeding up: $0\leq t<\frac{13}{3}$ and $t>\frac{26}{3}$
• Slowing down: $\frac{13}{3}<t<\frac{26}{3}$

For $y = 2x^{\frac{2}{3}}+3x^{-3}+5x^{-5}$, find the fourth - derivative.

Step6: Find the first - derivative

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, $y^\prime=\frac{d}{dx}(2x^{\frac{2}{3}}+3x^{-3}+5x^{-5})=2\times\frac{2}{3}x^{-\frac{1}{3}}-9x^{-4}-25x^{-6}=\frac{4}{3}x^{-\frac{1}{3}}-9x^{-4}-25x^{-6}$

Step7: Find the second - derivative

$y^{\prime\prime}=\frac{d}{dx}(\frac{4}{3}x^{-\frac{1}{3}}-9x^{-4}-25x^{-6})=\frac{4}{3}\times(-\frac{1}{3})x^{-\frac{4}{3}}+36x^{-5}+150x^{-7}=-\frac{4}{9}x^{-\frac{4}{3}}+36x^{-5}+150x^{-7}$

Step8: Find the third - derivative

$y^{\prime\prime\prime}=\frac{d}{dx}(-\frac{4}{9}x^{-\frac{4}{3}}+36x^{-5}+150x^{-7})=-\frac{4}{9}\times(-\frac{4}{3})x^{-\frac{7}{3}}-180x^{-6}-1050x^{-8}=\frac{16}{27}x^{-\frac{7}{3}}-180x^{-6}-1050x^{-8}$

Step9: Find the fourth - derivative

$y^{(4)}=\frac{d}{dx}(\frac{16}{27}x^{-\frac{7}{3}}-180x^{-6}-1050x^{-8})=\frac{16}{27}\times(-\frac{7}{3})x^{-\frac{10}{3}} + 1080x^{-7}+8400x^{-9}=-\frac{112}{81}x^{-\frac{10}{3}}+1080x^{-7}+8400x^{-9}$

Answer:

1. The particle in $s(t)=t^{3}-23t^{2}+120t$ changes directions at $t=\frac{10}{3}$ and $t = 12$.
2. For $s(t)=t^{3}-13t^{2}$, the particle is speeding up on the intervals $[0,\frac{13}{3})\cup(\frac{26}{3},\infty)$ and slowing down on the interval $(\frac{13}{3},\frac{26}{3})$.
3. For $y = 2\sqrt[3]{x^{2}}+3x^{-3}+\frac{5}{x^{5}}$, $\frac{d^{4}y}{dx^{4}}=-\frac{112}{81}x^{-\frac{10}{3}}+1080x^{-7}+8400x^{-9}$