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Castiel

  • one year ago

Find points on xyz-e^x+y^2=3 where tangent planes in those points also pass through T(2,0,1) and are parallel to z axis

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  1. Castiel
    • one year ago
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    Ok i solved for gradient \[grad=(y z-e ^{x})i+(xz+2y)j+(xz)k\]

  2. Castiel
    • one year ago
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    Im having problem figuring out the condition for the plane being parallel to z axis, so far I've come up with grad*(0,0,1)=0 as scalar product. From that I get xy=0

  3. Castiel
    • one year ago
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    My bad, the grad is actually grad=(yz−ex)i+(xz+2y)j+(xy)k, the last part is not xz but xy

  4. madhu.mukherjee.946
    • one year ago
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    @Vocaloid can you pls help

  5. Vocaloid
    • one year ago
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    it's been a while since I've done a problem like this @phi @ganeshi8

  6. madhu.mukherjee.946
    • one year ago
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    @phi

  7. madhu.mukherjee.946
    • one year ago
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    @ganeshie8

  8. madhu.mukherjee.946
    • one year ago
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    @LynFran

  9. Castiel
    • one year ago
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    I'm thinking from xy=0 we know that either x can be 0 or y, or both? So the points would be T1(x1,0,z1), T2(0,y2,z2) and T3(0,0,z3), maybe I'm wrong idk

  10. phi
    • one year ago
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    plug in (2,0,1) into the gradient

  11. ganeshie8
    • one year ago
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    (2, 0, 1) need not exist on the given surface right

  12. phi
    • one year ago
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    yes, that was the wrong thought

  13. Castiel
    • one year ago
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    no it does not, when I plug 2,0,1 in grad I get (-e^2)i+2j

  14. ganeshie8
    • one year ago
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    so the tangent plane looks like \[(y_0z_0−e^{x_0})x+(x_0z_0+2y_0)y = d\]

  15. Castiel
    • one year ago
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    yes, and d would equal to 2(y0z0-e^x0)

  16. ganeshie8
    • one year ago
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    so the tangent plane looks like \[(y_0z_0−e^{x_0})(x-2)+(x_0z_0+2y_0)y = 0\]

  17. ganeshie8
    • one year ago
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    Yes..

  18. phi
    • one year ago
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    parallel to z axis means the normal has the form ai + bj + 0k this means the 3rd component of the gradient is 0 xy = 0 so either x =0 or y=0. maybe this idea is worth pursuing

  19. Castiel
    • one year ago
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    I'm trying to think of something to do with that. But I just can't seem see where I could use that T(2,0,1)

  20. phi
    • one year ago
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    \[ xyz-e^x+y^2=3 \] if we try y=0 we get \[ -e^x= 3 \\ e^x = -3\] and that is not possible. so it must be x=0

  21. phi
    • one year ago
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    with x=0, \[ xyz-e^x+y^2=3 \\ -1+y^2= 3 \\ y= \pm 2 \] so the points on the curve will be <0,2,z0> and <0,-2,z0>

  22. madhu.mukherjee.946
    • one year ago
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    @Castiel is your problem solved

  23. madhu.mukherjee.946
    • one year ago
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    @phi thank you so much for helping

  24. phi
    • one year ago
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    a point (0,2,a) (i.e. z= a ) will be on the curve xyz-e^x +y^2 = 3 the gradient at that point is <2a-1, 4, 0> that is the normal N to the plane N dot P = d where P is any point on the plane <2a-1,4,0> dot <0,2,a> = 8 thus d= 8 and the equation of the plane is <2a-1, 4, 0> dot <x,y,z> = 8 we know point (2,0,1) is on the plane so we have <2a-1,4,0> dot (2,0,1)= 2(2a-1) which means 2(2a-1)= 8 and a= 5/2 the normal to the plane is <4,4,0> and the equation of the plane is <4,4,0> dot P = 8 or \[ <1,1,0>\cdot P = 2 \] we should now go back and try y=-2 to find the other solution

  25. phi
    • one year ago
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    Find points on xyz-e^x+y^2=3 where.... I am getting (0,2,5/2) and (0,-2,-5/2)

  26. Castiel
    • one year ago
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    isn't the dot product for two vectors? how can we do it with a vector and a dot

  27. phi
    • one year ago
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    yes, a dot product is between two vectors. But exactly what is the question? can you ask your question a different way?

  28. Castiel
    • one year ago
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    I don't understand this part N dot P = d where P is any point on the plane <2a-1,4,0> dot <0,2,a> = 8 thus d= 8 we know point (2,0,1) is on the plane so we have <2a-1,4,0> dot (2,0,1)= 2(2a-1) why is <2a-1,4,0> dot (2,0,1) == <2a-1,4,0> dot <0,2,a>

  29. phi
    • one year ago
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    I am using the "vector equation" for a plane perhaps you use \[ a (x-x_0) + b(y-y_0) +c (z-z_0) = 0 \] ?

  30. Castiel
    • one year ago
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    Ohhh yeah I see

  31. phi
    • one year ago
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    if so, we can write that as \[ ax + by + cz = a x_0 +b y_0 + c z_0 = d \] where a, b, c, x0,y0,z0 are fixed numbers and then write that as \[ < a,b,c> \cdot <x,y,z> = d \]

  32. Castiel
    • one year ago
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    Thank you!!! This makes perfect sense now. I can't thank you enough!

  33. phi
    • one year ago
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    and we can shorten that up to \[ \vec{N} \cdot \vec{P} = d \] where N is the normal vector, and P represents an arbitrary point <x,y,z>

  34. Castiel
    • one year ago
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    If you liked this one I'm certain I'll have more differential equation problems I can't solve in the upcoming week :D My exam is slowly approaching.

  35. Castiel
    • one year ago
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    But thank you! I've been strugling with this one the whole day.

  36. phi
    • one year ago
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    yes, it was a bit painful. (especially when I wrote the d/dy of y^2 as 2y but with my hand-writing, thought it was zy... that caused some grief)

  37. Castiel
    • one year ago
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    happened to me a bunch of times too. 2's and z's don't go well together.

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