Assume that x equals x (t )and y equals y (t ). Let y equals x cubed plus 4 and StartFraction dx Over dt EndFraction =2 when xequals3. Find StartFraction dy Over dt EndFraction when xequals3.
Accepted Solution
A:
Answer:dy/dt(x=3)=54Step-by-step explanation:We know that we can write: [tex]\frac{dy}{dt}=\frac{dy}{dx} *\frac{dx}{dt}[/tex]and so evaluate it as a product of functions, that is for a given value of x or t, we get that [tex]\frac{dy}{dt}(s)=\frac{dy}{dx} (s) * \frac{dx}{dt}(s)[/tex].Now we are told that:[tex]1. x=x(t),\ 2.y=x^3+4,\ \frac{dx}{dt} =2,\ for \ x=3,[/tex] whenever it happens, this means the influence of t is hidden, and we only consider the result, that is, how much is x when the derivative has the value of 2, we are not concerned for what value of t it happens, since we get all the necessary information beforehand. Now we need to calculate [tex]\frac{dy}{dx}=3x^2 \rightarrow \frac{dy}{dx}(3)=3(3)^2=27[/tex].Therefore [tex]\frac{dy}{dt}(x=3)=\frac{dy}{dx} (x=3) * \frac{dx}{dt}(x=3)[/tex] or[tex]\frac{dy}{dt}(x=3)=27 *2 = 54[/tex].