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 September 24th, 2010, 10:02 AM #1 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Need a little guidance, derivative etc I've tried solving the following example, I think I'm on the right track but I need reassurance and help simplifying the result, because my ever so strict professor picks random people to show their solution at the blackboard, and I absolutely dread this. Here it is: You have the function f(x,y,z)= x^2+2yz+y^2 Imagine that you travel in the xyz-room among the following lines x(t)= 0.5 - (sqrt(3)/2)t y(t)=sqrt(3)/2+0.5t z(t)= 1 During your travels you continously get the values of the function f. a) State the function that shows how the functionvalues of f depend on t I think that means I should put the equations of x(t), y(t) and z(t) into f(x,y,z)= x^2+2yz+y^2 This gives me: t^2+ t+ 1+ sqrt(3) In the next question he asks how fast the function value increases/decreases when we pass the point that is prevailing when t=0 So, when t = 0: x= 0.5, y=sqrt(3)/2, z=1 It seems like they're looking for the "direction-derivative", how do I find the answer to how fast the function value increases/decreases when we pass the point that is prevailing when t=0..? I would really appreciate a little help!
 September 24th, 2010, 10:14 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Need a little guidance, derivative etc To find "how fast the function value increases/decreases when we pass the point that is prevailing when t=0" wouldn't you just use f(t) to find f'(0)? f'(t) = 2t + 1 f'(0) = 1
 September 24th, 2010, 10:15 AM #3 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Re: Need a little guidance, derivative etc Ok, makes sense. Thanks..but what does this tell us? That it increases with speed 1?
 September 24th, 2010, 10:22 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Need a little guidance, derivative etc It tells us the f(t)'s instantaneous rate of change with respect to t at t = 0 is one unit. Since the question only asked about the rate of change of f, this should suffice.
 September 24th, 2010, 10:27 AM #5 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Re: Need a little guidance, derivative etc Ok, cheers. However the prof states: "You will have to calculate the direction-derivative of the function f in a certain point and in a certain direction that corresponds to t=0. What does he mean by direction, he wants me to indicate and specify the point and direction....?? Im lost.. (and then to calculate the direction derivative with help from the gradient to f)
 September 24th, 2010, 10:32 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Need a little guidance, derivative etc Hmmm. If your prof. wants the direction-derivative, I believe you need to compute the partials for f with respect to x,y,z. It has been a long time since I have done this, so let me review some and get back to you.
 September 24th, 2010, 10:40 AM #7 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Re: Need a little guidance, derivative etc Thanks so much for your help, I've done the partial derivatives and I get: 2x, 2z+2y, 2y What now... I think the prof wants me to understand the direction only with the help from the t-function in the first question, because it is only later that he starts mentioning the gradient, in which I calculate the directionderivative with the help from partial derivatives..
 September 24th, 2010, 02:10 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Need a little guidance, derivative etc Ok, I may be way off here, but... It seems we need to find the directional derivative of f(x,y,z) = x² + 2yz + y² in the direction of: [x(0)]i + [y(0)]j + [z(0)]k = (½)i + [½?(3)]j + k Since ?(½)i + [½?(3)]j + k? = ?(2), u = [?(1/]i + [?(3/]j + [?(1/2)]k We have ?f/?x = 2x, ?f/?y = 2z + 2y, ?f/?z = 2y Vf(x,y,z) = (2x)i + (2z + 2y)j + (2y)k Vf(½,½?(3),1) = (1)i + (2 + ?(3))j + (?(3))k It follows then that: Duf(½,½?(3),1) = [(1)i + (2 + ?(3))j + (?(3))k]·[[?(1/]i + [?(3/]j + [?(1/2)]k] = {2?(2)[1 + 2?(3)] + ?(3)[2 + ?(3)]}/8 ? 1.52 I was somehow expecting it to agree with the value of f'(0) = 1. Therefore, I will wait until someone who knows more about this can clear up this mess I've made.
 September 25th, 2010, 08:00 AM #9 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Re: Need a little guidance, derivative etc Thank you, it cleared up a little..it's a very hard task in my opinion.
 September 25th, 2010, 09:30 AM #10 Member   Joined: Sep 2010 Posts: 32 Thanks: 0 Re: Need a little guidance, derivative etc One more thing MarkFL, once you know that Duf(½,½?(3),1) = (1, 2+sqrt(3), sqrt(3)), how do you find the one value of the derivative, I don't understand the last part: [(1)i + (2 + ?(3))j + (?(3))k]·[color=#FF4000][[?(1/]i + [?(3/]j + [?(1/2)]k][/color] What formula are you using? My prof didn't provide me with one.

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