Question

For this question ( http://f.imgtmp.com/ghnor.jpg ), I'm trying to get either the gradient of both things or the tangent of both things and then dotting them together.

My problem is that I when I try to find the gradient of the curve, I do not know how to get z = f(x,y) for it so that I can partial differentiate that for each component of the gradient.

Similarly, if if I try to use tangents instead, I can't figure out how to get the surface with t variables such that I can find the tangent to it. If I am being unclear, just ask me to clarify even if you're not planning on helping out.

Answer 1

You can get a g(x,y,z) = 0 and derivate it to get the components of the gradient

Answer 2

for g(x,y,z) = 0, are you talking about the curve or the surface?

Answer 3

the surface

Answer 4

i already knew that as i explained in the first post (i probably didn't explain well though sorry). my problem is for the curve with the t variables. what do i do with the t variables? how do I make them x, y and z such that i can compute the gradient?

Answer 5

could you show me the steps? or tell me what to do step by step and i'll to the tedious algebra for you.

Answer 6

let me see my dropbox

Answer 7

alright. (does that have anything to do with me or you're just basically telling me to wait?)

Answer 8

take the derivative of the paramaterized curve w/respect to t and use one of the points to solve for t

Answer 9

one of the coordinates*

Answer 10

so you're doing it the tangents way?

Answer 11

let the curve be\[r(t)=<\frac23(t^3+2),2t^2,3t-2>\]the tangent will be given by the derivative\[r'(t)=<2t^2,4t,3>\]at the point we can use any coordinate to find t. Let's use the y-coordiante\[2t^2=2\to t=1\]so we have the direction of our curve at that point given by\[r'(1)=<2,4,3>\]

Answer 12

ok so r'(1) = <2,4,3> = tangent at point (2,2,1) foor the curve, right?

Answer 13

i likely am asking stupid stuff but i am trying to think in this sleep deprived state that i'm in

Answer 14

the gradient of the surface will be a vector perpendicular to that surface\[\nabla f=<2x,4y,6z>\]the vector at that point is\[\nabla f(2,2,1)=<4,8,6>\]if we take the dot product and compare it with the magnitudes we find that\[\nabla f(2,2,1)\cdot r'(1)=|\nabla f(2,2,1)||r'(1)|=58\]which means the angle between those vectors is zero, which proves what we wanted

Answer 15

yes to your question

Answer 16

how does the = 58 mean the angle is 0?

Answer 17

in the formula for the dot product:\[a\cdot b=|a||b|\cos\theta\]when the angle is 0 we have\[a\cdot b=|a||b|\cos0=|a||b|(1)=|a||b|\]so if the dot product of two vectors is the same as the product of their magnitudes, they are parallel

Answer 18

in this case we have\[\nabla f\cdot r'=|\nabla f||r'|\]\[<2,4,3>\cdot<4,8,6>=(\sqrt{2^2+4^2+3^2})(\sqrt{4^2+8^2+6^2})=58\]so that proves they are parallel

Answer 19

yes, i get it :D basically you looked for (dot product) = (just the mangnitude multiplications) because the question implied to look for it by asking you to prove it?

Answer 20

using the formula\[a\cdot b=|a||b|\cos\theta\]is a common way to find if vectors are parallel, perpendicular, or neither.

I used it to prove that the tangent vector is parallel to and coincides with the gradient vector, which means it must be perpendicular to the surface.

Answer 21

so yes, basically

Answer 22

for (b), why are there to equations for the curve and one for the surface?

Answer 23

when two vectors are paralell we have\[a\cdot b=|a||b|\]and when they are perpendicular we have\[a\cdot b=0\]

Answer 24

Yes I was wondering the same about b...
between that and the 'pint' typo I don't think I can help you with that part much

Answer 25

so it should be either x^2 - y^2 + z^2 = 1 OR xy + xz = 2 for the curve?

Answer 26

or maybe it's asking to verify for two curves and one surface?

Answer 27

I'm trying for the first curve but it's not looking right :/

Answer 28

maybe the second one is the intended one?

Answer 29

neither of the first two formulas are curves, they are both surfaces...
perhaps the 'curve' is the intersection of those surfaces...?

Answer 30

that sounds like a very difficult problem, and I don't know if I want to undertake it without being sure that's what's requested

the wording is clearly very strange :/

Answer 31

I think I have an idea...

Answer 32

I think they want to prove that each surface individually is tangent to the surface xyx-x^2-6y+6=0

Answer 33

so we need first to notice that all those surfaces share the point (1,1,1), so they at least intersect in some manner through that point

to prove they are tangent we need to prove their gradients are parallel, which we would do with the same formula we just used

Answer 34

would you be willing to be my math slave in exchange for me making you a graphical program for whatever it is you want? :P

Answer 35

let g(x,y,z)=x^2 - y^2 + z^2 -1
and h(x,y,z)=xy + xz = 2
and f(x,y,z)= xyx-x^2-6y+6

we have to prove that\[\nabla g\cdot\nabla f=|\nabla g||\nabla f|\]and \[\nabla h\cdot\nabla f=|\nabla h||\nabla f|\]at the point in questoin

and no thanks, teaching just helps me practice ;)

Answer 36

lol damn was just blowing off the cob webs from my basement's chains

Answer 37

would you be willing to do another 3 problems while i sleep? lol

Answer 38

I don't think the results would be up to par if I don't get my sleep

Answer 39

o ii didn't think your time zone was like mine

Answer 40

i would nevewr ask anyone to lack sleep lol

Answer 41

did you completely answer (b)?

Answer 42

well not the tedious algebra but the logic

Answer 43

no I'm still not getting a decisive answer, though my current interpretation is all I can come up with

Answer 44

I mean, I'm just not sure what the damn thing is asking

Answer 45

i'll look it over again tomorrow morning based on what you attempted. i don't want to keep you up for this. i will sleep and so should you. i appreciate your help very much.

Answer 46

I would be more persistent with it, but with the fuzzy wording I can't continue much

you're welcome, good luck :D

Answer 47

thanks again :)

Answer 48

ttyl