Question

Assuming that the equation defines x and y implicitly as differentiable functions xequals=​f(t), yequals=​g(t), find the slope of the curve xequals=​f(t), yequals=​g(t) at the given value of t. x^3+3t^2 =13, 2y^3−3t^2= 42, t=2. The slope of the curve at t=2 is what?

Answer 1

Answer:

$\frac{-1}{18}$

Step-by-step explanation:

Given that x and y are implicitly as differentiable functions.

xequals=​f(t), yequals=​g(t),

$x^3+3t^2 =13,$

$2y^3-3t^2= 42$

we have to get value of x and y at t =2

$x^3+3(4) = 13\\x =1\\2y^3-12 = 42\\y^3 = 27\\y=3$

we have to find the slope of the curve at t=2

i.e. we have to find $\frac{dy}{dx}$

=$\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$ at t=2

$x^3+3t^2 =13\\3x^2 \frac{dx}{dt} +6t = 0$$2y^3-3t^2= 42\\6y^2 \frac{dy}{dt} -6t = 0$

Substitute the values of x and y and also t in these equations to get

$3(1)^2 \frac{dx}{dt} +6(2) = 0\\\frac{dx}{dt} =-4\\6(3)^2 \frac{dy}{dt} -6(2)= 0\\\frac{dy}{dt}=\frac{2}{9}$

Slope = $\frac{2/9}{-4}=\frac{-1}{18}$