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

Answer 1

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

Answer 2

Derivative of ln(u) = du/u

Answer 3

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

Answer 4

Derivative of ln(2t) = 2/(2t) = 1/t

So the derivative is 3(1/t) = 3/t

Answer 5

Oh i see, so what do I do next?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

so dy/dx = 25 right?