A curve has parametric equations, where t is a positive parameter.

Question

The parametric equation is {x = e ^√ t , y =t - ln (t^2) ,

a) Find the dy/dx and d^2y/dx^2 in terms of t.

b) Find the equation of the tangent line at the point when t =1.

Answer 1

a) dx/dt=(e^√t)/(2√t), y=t-2ln(t), dy/dt=1-2/t=(t-2)/t.

dy/dx=(dy/dt)/(dx/dt)=((t-2)/t)(2√te^-√t)=2(t-2)(√t/t)e^-√t. √t/t can also be written 1/√t and dy/dx can be written: dy/dx=2(t-2)/(√te^√t).

Let p=dx/dt and q=dy/dt, dy/dx=q/p and d²y/dx²=d(q/p)/dx=(d(q/p)/dt)/p.

So d²y/dx²=(pdq/dt-qdp/dt)/p³.

dp/dt=d²x/dt²=(e^√t/4t)(1-1/√t), dq/dt=d²y/dt²=2/t², pdq/dt=e^√t/(t²√t),

-qdp/dt=-(1-2/t)(e^√t/4t)(1-1/√t)=(e^√t/(4t³))(t√t+2t-2√t-t²) after rationalisation,

p³=e^(3√t)/(8t√t).

Therefore, d²y/dx²=(e^√t/(t²√t)+(e^√t/(4t³))(t√t+2t-2√t-t²))/(e^(3√t)/(8t√t)).

That is, d²y/dx²=(2√t/(t²e^(2√t))(4+t√t+2t-2√t-t²) (I think!)

b) When t=1, dy/dx=2(t-2)(1/√t)e^-√t=-2/e, which is the gradient at t=1. The equation of the tangent is of the form y=-2x/e+a where a is a constant. (x,y)=(e^t,t-ln(t²))=(e,1), so 1=-2+a, a=3. Therefore y=-2x/e+3 is the tangent line when t=1.