Chain Rule for Related Rates

We are given that

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

Using the chain rule,

$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$

Therefore,

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

Assuming the particle is moving,

$$\frac{dx}{dt}\neq0$$

Hence,

$$\frac{dy}{dx}=2$$

Now differentiate the curve:

$$y=\frac13x^3-3x^2+11x-5$$

$$\frac{dy}{dx}=x^2-6x+11$$

Set this equal to

$$2$$

$$x^2-6x+11=2$$

$$x^2-6x+9=0$$

$$(x-3)^2=0$$

$$x=3$$

Substitute into the equation of the curve:

$$y=\frac13(3)^3-3(3)^2+11(3)-5$$

$$=9-27+33-5$$

$$=10$$

Therefore the required point is

$$(3,10)$$

Final Answer :

$$(3,10)$$