AboutAlon 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".
Question 1. Find the equation of the normal to the curve y = 2x / x^2 + 1 at the point (3;0,6) and the equation of the tanget at the origin.
2. If y = sec x, prove that y d^2y/ dx^2 = (dy/dx)^2 + y^4.
3. Determine the maximum and minimum points and sketch the graph of y = 6x^4 - 8x^3 + 1.
4. Determine the area enclosed by the following boundaries y = x^2 - 4x + 20, y = 3x, x = 0 and x = 4.
5. The number of radio active atom particles (N) is given as a function of time elapsed (t), that is: N = 10e^-t/50
Find the rate of decline in N.
Answer 1. Let's start with the tangent :
The tangent line has slope of 'derivative' of the function in
the specific point. Meaning, if the equation of the tangent is
y=ax+b, then a=f'(x) at x=0. Where f(x)=2x/x^2+1.
After calculating a, we need to calculate b.Which is the point
intersection with the y-axis ; which is zero. So the equation is
y=ax.
The normal line : the normal's slope is -1/(tangent's slope)
because it's perpendicular to it. So 1st find the slope of the
tangent at (3,0.6) & then calculate the slope of the normal &
the intersection point with y-axis.
2. Her's what we know :
1) y=sec(x)
2) derivative of sec(x)=sec(x)tang(x).
3) derivative of tang(x)=[sec(x)]^2.
Let's get to work:
y'=sec(x)tan(x)
y''=sec(x)tang(x)tang(x)+sec(x)[sec(x)]^2
y''=sec(x)[tang(x)]^2+[sec(x)]^3
y*y''=[sec(x)]^2*[tang(x)]^2+[sec(x)]^4
y*y''=[ sec(x)tang(x) ]^2 + [sec(x)]^4
y*y''=[y']^2+y^4.
Q.E.D
3. To find the extremum points you need to derive the function &
set it equal to zero. 24x^3-24x^2=0. x=0,x=1.
y''=72x^2-48x. y''(0)=0. y''(1)>0 so,
(x=0) is a saddle point, (x=1) is minima
4. Area = Integral(x from 0 to 4) of { [x^2-4x+20]-[3x] }
the bounded area is the area between the two graphs.
Area = Integral(x from 0 to 4) of { x^2-7x+20 }
5. Rate of decline is the derivative
N'=(-1/50)*10*e^(-t/50)
N'=(-1/50)N
