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.

in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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.

 

by Top Rated User (652k points)
reshown by

Related questions

1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,035 questions
87,752 answers
1,973 comments
4,581 users