16) determine when the object changes direction according to the parametric equations on t = 0, 6. select…

16) determine when the object changes direction according to the parametric equations on t = 0, 6. select all t - values that apply.\n$x(t)=\frac{1}{3}t^{3}-9t,y(t)=4t^{2}-16t$\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nnever

16) determine when the object changes direction according to the parametric equations on t = 0, 6. select all t - values that apply.\n$x(t)=\frac{1}{3}t^{3}-9t,y(t)=4t^{2}-16t$\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\nnever

Answer

Explanation:

Step1: Find the velocity - functions

The velocity functions are the derivatives of the position functions. $x'(t)=\frac{d}{dt}(\frac{1}{3}t^{3}-9t)=t^{2}-9$ $y'(t)=\frac{d}{dt}(4t^{2}-16t)=8t - 16$

Step2: Determine when the velocity - components change sign

For the $x$ - component: Set $x'(t)=t^{2}-9 = 0$. Then $(t - 3)(t + 3)=0$. The solutions are $t = 3$ and $t=-3$. In the interval $t\in[0,6]$, we consider $t = 3$. When $0\leq t<3$, $x'(t)=t^{2}-9<0$. When $3<t\leq6$, $x'(t)=t^{2}-9>0$. For the $y$ - component: Set $y'(t)=8t - 16=0$. Solving for $t$ gives $t = 2$. When $0\leq t<2$, $y'(t)=8t - 16<0$. When $2<t\leq6$, $y'(t)=8t - 16>0$.

Answer:

2, 3