The parametric equations of the curve are `x=2t(t^2+3)-3t^2, y=2t(t^2+3)+3t^2` Find the maximum slope of the curve and the corresponding point on it.

Parametric equations a line are x=2-3t, y=-1+2t and z=3+t. write down the vector equations of the line.

Find the slope of the tangent to the curve y=sin 3t,x=2t at t=pi/4

For the curve x=t^(2) +3t -8 ,y=2t^(2)-2t -5 at point (2,-1)

For the curve x=t^(2) +3t -8 ,y=2t^(2)-2t -5 at point (2,-1)

Consider the curve represented parametrically by the equation x=t^(3)-4t^(2)-3t and y=2t^(2)+3t-5 where t in R. If H denotes the number of point on the curve where the tangent is horizontal and V the number of point where the tangent is vertical then

Normal equation to the curve y^(2)=x^3 at the point (t^(2),t^(3)) on the curve is

Find the equation of tangent to the curve x=sin 3t,y=cos 2t at t =pi/4

Find the slope of the tangent to the curve x=t^2+3t-8 , y=2t^2-2t-5 at t=2 .

Find the slope of the tangent to the curve x=t^2+3t-8 , y=2t^2-2t-5 at t=2 .