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Expert: Alon Mandes - 2/5/2009


Question
A point is moving along the curve y=6e to the power of X squared over 4 in such a way that the horizontal component dx/dt of its velocity is always 3. Find dy/dt at the instant when X=2.

Answer
our curve is y=f(x). That means : dy/dt=(dy/dx)(dx/dt).We know that
dx/dt=3 & dy/dx=y'(x). Thus,
dy/dt {at x=2} = 3*y'(2). Let's 1st calculate the derivative :
y=6e^[(x^2)/4]
y'=[x^(2)/4]*[(2/4)x]*6e^[(x^2)/4]
y'=(3/4)(x^3)e^[(x^2)/4]
y'(2)=(3/4)e^[8/4]=5.541
So,dy/dt {at x=2} = 3*5.541 = 16.62

Alon.

Calculus

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Alon Mandes

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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".

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

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

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