Question

a point is moving along the curve y=2x^3-3x^2 in such a way that it's x-coordinate 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)

Answer 1

distance from origin , P^2 = x^2 + y^2
we have to find dP/dt which is = 2x dx/dt + 2y dy/dt
= 2 (2) dx/dt + 2(4)dy/dt
= 4dx/dt + 8dy/dt

Now dx/dt = 2 cm/sec
dy/dt = 6x^2 - 6x
AT (2,4) dy/dt = 24-12 = 12cm/sec

therefore dP/dt = 4*2 + 8*12
= 104

Answer 2

thanks for help but i think there are some mistakes

Answer 3

akshat, you forgot to take sqrt of P^2

P = sqrt(x^2 +y^2)
dP/dx = x/sqrt(x^2 +y^2) --> dP/dx = 1/sqrt5
dP/dy = y/sqrt(x^2 +y^2) --> dP/dy = 2/sqrt5
dy/dx = 6x^2 -6x ---> dy/dx = 12
\[\frac{dP}{dt} = \frac{dx}{dt}*\frac{dP}{dx} +\frac{dx}{dt}*\frac{dy}{dx}*\frac{dP}{dy}\]
\[\frac{dP}{dt} = 2*\frac{1}{\sqrt{5}}+2*12*\frac{2}{\sqrt{5}} = \frac{50}{\sqrt{5}} = 10\sqrt{5}\]

Answer 4

Oh yes...sorry!

Answer 5

You can do it this way (minor variation on dumbcow)
\[ P= \sqrt{x^2+y^2} \]
Taking the derivative wrt to t
\[ P' = \frac{ x x' + y y'}{\sqrt{x^2+y^2}} \]
where P', x', y' are the derivative with respect to t
x'= 2 (given)
y' = (6x^2-6x) x'
At (2,4), with x'=2, y'= 24
P'= (2*2+4*24)/sqrt(20)= 100/sqrt(20) = 50/sqrt(5)= 10 sqrt(5) cm/sec