A curve is defined by the parameters x =3ln(2t) and y = t^3 - 2t

Find the gradient of the tangent to the curve at t = 3

Show any derivatives that you need to find when solving this problem

Question

A curve is defined by the parameters x =3ln(2t) and y = t^3 - 2t

Find the gradient of the tangent to the curve at t = 3

Show any derivatives that you need to find when solving this problem

anonymous · Answer

I know that dy/dt = 3t^2 - 2 but idk how to differentiate 3ln(2t)

anonymous · Answer

Derivative of ln(u) = du/u

anonymous · Answer

so how does 3ln(2t) differentiate into 3/t?

anonymous · Answer

Derivative of ln(2t) = 2/(2t) = 1/t

So the derivative is 3(1/t) = 3/t

anonymous · Answer

Oh i see, so what do I do next?

anonymous · Answer

So now, you have dy/dx, now plug in t=3

anonymous · Answer

because dy/dx = (dy/dt)/(dx/dt)

anonymous · Answer

dy/dx is an expression in t, since you have parametric functions.

anonymous · Answer

so dy/dx = 25 right?