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Homework Help: Need help with some Mutilvariable problems

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data
    We just started Multivariable at school after AP and I have a problem set to do. Since, I am newbie, so I am struggling a bit.

    1. Suppose the current I in an electrical circuit is related to voltage V and the resistance R by the equation I=V/R. If voltage drops from 24 to 23 volts and resistance drops from 100 ohms to 80 ohms, will I increase or decrease by how much ?

    2. Find the derivation of f(x,y,z)=xyz in the direction of the velocity vector of the helix r(t)=cos3t i + sin3t j + 3t k at t=pi/3

    3. At the point (1,2) the function f(x,y) has a derivative of 2 in the direction toward (2,2) and derivative of -2 in the direction toward (1,1)

    a. Find D1f(1,2) and D2f(1,2)
    b. Find the derivative of f at (1,2) in the direction toward (4,6)

    2. Relevant equations


    3. The attempt at a solution

    1. So I took the partial derivative and got D1f= 1/R and D2f= -V/R^2.

    2. So I found the gradient vector for f as yz i + xz j + xy k

    I also found r(pi/3)= -1 i + pi k

    How do I find the value of x,y,z and also, do I need to find the unit vector of r ?

    So for delta I = 1/R*delta V - V/R^2*delta R. Plugging in the numbers I got a decrease but it should be an increase. Where did I do wrong ?

    3. Any suggestion on how to do this problem ??
  2. jcsd
  3. May 24, 2009 #2
    So for number 2, do I need to take the unit vector or just multiply gradient by r ?
  4. May 24, 2009 #3
    Hi, the derivative of a function in the direction of a vector is just a scalar product of the (unit) vector with the gradient of the function.
    1) get the point P(x0,y0,z0) by plugging pi/3 for t into the parametric equation of the helix. P=(-1,0,pi)
    2) get the velocity vector of the helix: differentiate each component of r(t) with respect to t: r´(t) = (-3*sin3t, 3*cos3t, 3)
    now we plug again pi/3 for t and we get (0,-3,3)
    we compute the unit velocity vector (0, -3/sqrt(18), 3/sqrt(18)), at the point (-1,0,pi)
    3) make a scalar product of unit r´(t) with Grad(f) and plug -1,0,pi and you get 0+((-1)*pi*-3/sqrt(18)) +0
    4) the result is 3*pi/sqrt(18)

    or read this http://tutorial.math.lamar.edu/pdf/CalcIII/CalcIII_DirectionalDerivatives.pdf

    PS: sorry I got it wrong the first time
    Last edited: May 24, 2009
  5. May 24, 2009 #4

    That's what I got. However, anyway that I can find out what is x,y,z ??
  6. May 25, 2009 #5
    Thanks. How about other problems ?
  7. May 25, 2009 #6
    What is the total derivative of I for part a)? You found some of the parts for it, the partial derivatives. Now put them into the form dI = something. The usual rule for these type of estimation problems is then to change the d's to delta's. Plug in the numbers for the delta's.
  8. May 25, 2009 #7
    So is it delta I= 1/R*delta V - V/R^2*delta R ?

    When I plug in the number, I got decreasing but should it be increasing ?
  9. May 25, 2009 #8
    Your formula looks good. Try the math again, it should work. Remember Delta = final value - initial value. So both your delta's should be negative when you plug them in.
  10. May 25, 2009 #9
    If I just plug in V and R into I=V/R then I have an increasing (24/100 is smaller than 23/80 right ?) but if I plug them in the differentiation that I just got, I got a decreasing.
  11. May 25, 2009 #10
    write out how the numbers you are putting in each term. I get a positive or increasing value.
  12. May 25, 2009 #11
    so I plugged in 80 for R, 1 for delta v, 23 for V and 20 delta R. Thus I got 1/80- (23*20/80^2) and a negative result
  13. May 25, 2009 #12
    read what I said above.

    Delta = final value - initial value. Both your deltas should be plugged in as negative numbers.

    For example:
    Delta V = final v - initial v = 23-24 = -1

    Delta R will also be negative.
  14. May 25, 2009 #13
    my bad, did you get approximately 0. 06 ? Thanks

    How about the last problem ?
  15. May 25, 2009 #14
    If I use those numbers then yes that's about it. The only comment I have is I can't tell you for sure what you should plug in for V and R. If you use the initial values of 24 and 100, instead of 23 and 80 you get a more accurate number. If I was doing this in real life, I'd probably use the average of the two number 23.5 and 90. You'll just have to look for an example that your text or teacher did to see what convention they use.
  16. May 25, 2009 #15
    Thanks. How problem number 3, can you give me a hint on how I should do it ?
  17. May 25, 2009 #16
    I'm not totally sure, however, it seems to be a reverse of what we just did. Write out the general form for the total differential. Change the d's to delta's. The problem statement gives you three different points to work with. Compute a Delta X and a Delta Y for two of them on one side of this equation and the Delta on the other side of the equation is given to you. This gives you one equation with D1F and D2F as the unknowns. Now do the same thing for the other two points to get a second equation. Solve for D1F and D2F. Then use them to find the differential at the third direction.
  18. May 25, 2009 #17
    ...actually it turns out to be simpler because Delta X is zero for one set of points and Delta Y is zero for the other set of points. So you get the unknown D1f and D2f without having to solve two simultaneous equations.
  19. May 25, 2009 #18
    So just use directional derivative I got:

    D1f*1 +0=2 and 0+D2f*(-1)=-2

    Thus, I got D1f= 2=D2f. Does it sound right ?

    for part b, I got 14/5 after scalar D1f and D2f with unit vector 3/5 i +4/5 j. Sound right ?
  20. May 25, 2009 #19
    Actually, I got it wrong the first time. I was confusing total differentials with directional derivatives. (It's been too long :) After seeing your answer, I did a quick refresher of my text and your answer looks correct to me. So we both learned something.
  21. May 25, 2009 #20
    Thanks. Yeah, I am new too this since it's only my 2nd week of learning.

    Also a question, when I am asked for the parametric equations for the line that is normal to the tangent plane at a point, do I just take D1f, D2f, D3f and calculate the value at the point then plug in parametric equation like

    x=D1f*t +x0
    y=D2f*t +y0
    z=D3f*t +z0
  22. May 25, 2009 #21
    The gradient of the function that defines the surface points in the normal direction. So your D1f, D2f, D3f from the gradient are components of the normal vector. So the parametric line equations should be as you gave them.
  23. May 25, 2009 #22
    Thank you
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