Answer:
(0,1) and (13,2)
Step-by-step explanation:
We are given that parametric equations
[tex]x=2t^3+3t^2-12t[/tex]
[tex]y=2t^3+3t^2+1[/tex]
a.We have to find the points where tangent line is horizontal.
Differentiate x and y w.r.t.time
[tex]\frac{dx}{dt}=6t^2+6t-12[/tex]
[tex]\frac{dy}{dt}=6t^2+6t[/tex]
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Substitute the values
[tex]\frac{dy}{dx}=\frac{6t^2+6t}{6t^2+6t-12}[/tex]
We know that when tangent line is horizontal then
[tex]\frac{dy}{dx}=0[/tex]
[tex]\frac{6t^2+6t}{6t^2+6t-12}=0[/tex]
[tex]6t^2+6t=0[/tex]
[tex]t^2+t=0[/tex]
[tex]t(t+1)=0[/tex]
[tex]t=0, t+1=0[/tex]
[tex]t+1=0\implies t=-1[/tex]
Substitute the t=0 then we get
[tex]x=2(0)^3+3(0)^2-12(0)=0[/tex]
[tex]y=2(0)^3+3(0)^2+1=1[/tex]
Substitute t=-1
[tex]x=2(-1)^3+3(-1)^2-12(-1)=-2+3+12=13[/tex]
[tex]y=2(-1)^3+3(-1)^2+1=-2+3+1=2[/tex]
The points (0,1) and (13,2) where tangent line is horizontal .
