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About Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

Experience
1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

Education/Credentials
M.A in Mathematics & Bs.c in Electronics.

You are here: Experts > Teens > Homework/Study Tips > Calculus > tangent lines on a curve

Calculus - tangent lines on a curve

Expert: Alon Mandes - 11/4/2009

Question
Homework help calculus 2 is killing me...this question ask to find the equation of the tangent line for the curve represented by x= square root of t and y = 1/2t^2.
I started the problem by taking the derivative of dx/dt= which gave me t, and dy/dt= i got 1/square root of t/2 and from there i did (dy/dt)/(dx/dt) and got stuck from there because i didn't know where to go from there. Any help would be great thank you

Answer
The parametrization of the curve : x=√t & y=1/[2t²] . To find slope of the tangent line, we
need to find y'=dy/dx of the curve . We can do it in 2 ways :

1st Way :
dy/dt (1/[2t²])' -(1/2)*2*1/t³ 2√t 2x 2
dy/dx= ------- = ---------- = -------------- = - ------- = -------- = ----- .
dx/dt (√t)' 1/(2√t) t³ x^6 x^5

2nd Way : y=1/[2t²] --> y(x)=1/[2x^(4)]. Therefore : dy/dx=y'(x)=(1/2)4*1/x^5=2/x^5 .

We calculated the slope of the function which is also the slope of the tangent line.

Alon.

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