Question
A particle is moving along the curve y= 3 \sqrt{3 x + 4}. As the particle passes through the point (4, 12), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.
Answers
is the curve
y = 3/√(3x + 4) ??
if so, then the particle does not pass through your given point (4,12)
after you establish where your error is,
the method to solve the problem would be:
differentiate your equation with respect to t
your differential equation contains a dy/dt and a dx/dt term.
sub in dx/dt = 4 when x = ? and y = ? from the correct given point.
now the distance from the origin of a general point on the curve is
d^2 = x^2 + y^2
2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)
sub in all the stuff from above
y = 3/√(3x + 4) ??
if so, then the particle does not pass through your given point (4,12)
after you establish where your error is,
the method to solve the problem would be:
differentiate your equation with respect to t
your differential equation contains a dy/dt and a dx/dt term.
sub in dx/dt = 4 when x = ? and y = ? from the correct given point.
now the distance from the origin of a general point on the curve is
d^2 = x^2 + y^2
2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)
sub in all the stuff from above
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