Given:
[tex]\begin{gathered} x\left(t\right)=t^3-4t^2+t \\ y\left(t\right)=t^2-3t+3 \end{gathered}[/tex]
Required:
Find
[tex]x(3),y(3),\frac{dx}{dt},\text{ }\frac{dy}{dt},\frac{dy}{dx}\text{ and speed at t=3}[/tex]
Explanation:
The given functions are
[tex]\begin{gathered} x\left(t\right)=t^3-4t^2+t \\ y\left(t\right)=t^2-3t+3 \end{gathered}[/tex]
Substitute t=3 in x(t).
[tex]\begin{gathered} x\left(3\right)=(3)^3-4(3)^2+(3) \\ =27-36+3 \\ =-6 \end{gathered}[/tex]
Substitute t=3 in y(t).
[tex]\begin{gathered} y\left(3\right)=(3)^2-3(3)+3 \\ =9-9+3 \\ =3 \end{gathered}[/tex]
Differentiate the function x(t) with respect to t.
[tex]\frac{dx}{dt}=3t^2-8t+1[/tex]
Substitute t = 3 in
[tex]\frac{dx}{dt}[/tex][tex]\begin{gathered} \frac{dx}{dt}|_{t=3}=3(3)^2-8(3)+1 \\ =27-24+1 \\ =4 \end{gathered}[/tex]
Differentiate the function y(t) with respect to t.
[tex]\begin{gathered} \frac{dy}{dt}=2t-3 \\ \frac{dy}{dt}|_{t=3}=2(3)-3 \\ =6-3 \\ =3 \end{gathered}[/tex][tex]\begin{gathered} \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dy}{dx}} \\ \frac{dy}{dx}=\frac{2t-3}{3t^2-8t+1} \\ \frac{dy}{dx}|_{t=3}=\frac{2(3)-3}{3(3)^2-8(3)+1} \\ =\frac{3}{4} \end{gathered}[/tex]
The speed at 3 is 3/4.
F
