Answer:
Max at t=2, 128 m/s
Min at t=6, 0 m/s
Step-by-step explanation:
Given the position function of a particle with respect to time, find the minimum and maximum velocity the particle travels over the interval [1,7].
[tex]s(t)=t^4-16t^3+72t^2+5[/tex]
(1) - Find the velocity function of the particle
The velocity function is a derivative of the position function.
[tex]s'(t)=v(t)\\\\s(t)=t^4-16t^3+72t^2+5\\\\\Longrightarrow s'(t)=\frac{d}{dx}[t^4-16t^3+72t^2+5] \\\\\text{Use the derivative rules.}\\\\\boxed{\left\begin{array}{ccc}\text{\underline{Power Rule:}}\\\\\frac{d}{dx}[x^n]=nx^{n-1} \end{array}\right} \ \ \boxed{\left\begin{array}{ccc}\text{\underline{Constant Rule:}}\\\\\frac{d}{dx}[k]=0 \end{array}\right} \\\\\\\Longrightarrow s'(t)=(4)t^{4-1}-16(3)t^{3-1}+72(2)t^{2-1}+0\\\\\Longrightarrow s'(t)=4t^{3}-48t^{2}+144t\\\\[/tex]
[tex]\therefore \boxed{v(t)=4t^{3}-48t^{2}+144t}[/tex]
(3) - Take the derivative of v(t)
[tex]v(t)=4t^{3}-48t^{2}+144t\\\\\Longrightarrow v'(t)=12t^2-96t+144[/tex]
(4) - Let v'(t)=0 and solve for "t," these are the critical points
[tex]v'(t)=12t^2-96t+144\\\\\Longrightarrow 0=12t^2-96t+144\\\\\Longrightarrow 0=12[t^2-8t+12]\\\\\Longrightarrow 0=t^2-8t+12\\\\\Longrightarrow (t-6)(t-2)=0\\\\\therefore \text{The critical points are} \ \boxed{t=6 \ \text{and} \ t=2}[/tex]
(5) - Find the max/min values (in this case these values represent the particle's velocity) by plugging the critical points into v(t)
[tex]\text{Recall that} \ v(t)=4t^{3}-48t^{2}+144t \ \text{and} \ t=6, \ t=2\\\\\text{\underline{When t=6:}}\\\\\Longrightarrow v(6)=4(6)^{3}-48(6)^{2}+144(6)\\\\\Longrightarrow \boxed{v(6)=0 \ m/s}\\\\\text{\underline{When t=2:}}\\\\\Longrightarrow v(2)=4(2)^{3}-48(2)^{2}+144(2)\\\\\Longrightarrow \boxed{v(6)=128 \ m/s}[/tex]
Thus, at time, t=6, the particle's velocity is smallest, 0 m/s. And at time, t=2, the particle's velocity is greatest, 128 m/s.
