Answer:
[tex]\frac{-1}{18}[/tex]
Step-by-step explanation:
Given that x and y are implicitly as differentiable functions.
xequals=f(t), yequals=g(t),
[tex]x^3+3t^2 =13,[/tex]
[tex]2y^3-3t^2= 42[/tex]
we have to get value of x and y at t =2
[tex]x^3+3(4) = 13\\x =1\\2y^3-12 = 42\\y^3 = 27\\y=3[/tex]
we have to find the slope of the curve at t=2
i.e. we have to find [tex]\frac{dy}{dx}[/tex]
=[tex]\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex] at t=2
[tex]x^3+3t^2 =13\\3x^2 \frac{dx}{dt} +6t = 0\\[/tex][tex]2y^3-3t^2= 42\\6y^2 \frac{dy}{dt} -6t = 0\\[/tex]
Substitute the values of x and y and also t in these equations to get
[tex]3(1)^2 \frac{dx}{dt} +6(2) = 0\\\frac{dx}{dt} =-4\\6(3)^2 \frac{dy}{dt} -6(2)= 0\\\frac{dy}{dt}=\frac{2}{9}[/tex]
Slope = [tex]\frac{2/9}{-4}=\frac{-1}{18}[/tex]
