A point is moving along the curve $y=2x^3-3x^2$ in such a way that its $x$-cordinate is increasing at the rate of $2cm/sec$. Find the rate at which the distance of the point from the origin is increasing when the point is at $(2,4)$.
My Attempt: $y=2x^3-3x^2$ $$\dfrac {dy}{dt}=6x^2.\dfrac {dx}{dt}-6x.\dfrac {dx}{dt}$$ Also, $$\dfrac {dx}{dt}=2cm/sec$$ Now, the distance of the point $(x,y)$ from the origin is $$s^2=x^2+y^2$$ $$2s.\dfrac {ds}{dt}=2x\dfrac {dx}{dt}+2y.\dfrac {dy}{dt}$$ $$s.\dfrac {ds}{dt}=x.\dfrac {dx}{dt}+y.\dfrac {dy}{dt}$$ $$s.\dfrac {ds}{dt}=2\times 2+4\times (12x^2-12x)$$
How do I proceed further?
You can find $s$ and $(12x^2-12x)$ at the point $(2,4).$
Substitute the results in $$s.\dfrac {ds}{dt}=2\times 2+4\times (12x^2-12x)$$ and solve for $ \dfrac {ds}{dt}$