Answer:
[tex]\dfrac{d^2y}{dx^2}=0[/tex]
D is correct.
Step-by-step explanation:
Given: [tex]x(t)=2t+5[/tex]
[tex]y(t)=\dfrac{3t}{2}[/tex]
To find: [tex]\dfrac{d^2y}{dx^2}[/tex]
As we know,
[tex]\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
This is parametric equation. Differentiate both function separately and substitute into formula.
[tex]x(t)=2t+5[/tex] and [tex]y(t)=\dfrac{3t}{2}[/tex]
[tex]\dfrac{dx}{dt}=2,\dfrac{y}{dt}=\dfrac{3}{2}[/tex]
Substitute into derivative
[tex]\dfrac{dy}{dx}=\dfrac{3}{2\cdot 2}=\dfrac{3}{4}[/tex]
For double derivative differentiate w.r.t x
[tex]\dfrac{d^2y}{dx^2}=0[/tex]
Hence, The value of [tex]\dfrac{d^2y}{dx^2}=0[/tex]
