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Castiel
 one year ago
Find points on xyze^x+y^2=3 where tangent planes in those points also pass through T(2,0,1) and are parallel to z axis
Castiel
 one year ago
Find points on xyze^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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Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Ok i solved for gradient \[grad=(y ze ^{x})i+(xz+2y)j+(xz)k\]

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Im 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

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1My bad, the grad is actually grad=(yz−ex)i+(xz+2y)j+(xy)k, the last part is not xz but xy

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0@Vocaloid can you pls help

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0it's been a while since I've done a problem like this @phi @ganeshi8

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1I'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

phi
 one year ago
Best ResponseYou've already chosen the best response.3plug in (2,0,1) into the gradient

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0(2, 0, 1) need not exist on the given surface right

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, that was the wrong thought

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1no it does not, when I plug 2,0,1 in grad I get (e^2)i+2j

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so the tangent plane looks like \[(y_0z_0−e^{x_0})x+(x_0z_0+2y_0)y = d\]

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1yes, and d would equal to 2(y0z0e^x0)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so the tangent plane looks like \[(y_0z_0−e^{x_0})(x2)+(x_0z_0+2y_0)y = 0\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3parallel 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

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1I'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)

phi
 one year ago
Best ResponseYou've already chosen the best response.3\[ xyze^x+y^2=3 \] if we try y=0 we get \[ e^x= 3 \\ e^x = 3\] and that is not possible. so it must be x=0

phi
 one year ago
Best ResponseYou've already chosen the best response.3with x=0, \[ xyze^x+y^2=3 \\ 1+y^2= 3 \\ y= \pm 2 \] so the points on the curve will be <0,2,z0> and <0,2,z0>

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0@Castiel is your problem solved

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.0@phi thank you so much for helping

phi
 one year ago
Best ResponseYou've already chosen the best response.3a point (0,2,a) (i.e. z= a ) will be on the curve xyze^x +y^2 = 3 the gradient at that point is <2a1, 4, 0> that is the normal N to the plane N dot P = d where P is any point on the plane <2a1,4,0> dot <0,2,a> = 8 thus d= 8 and the equation of the plane is <2a1, 4, 0> dot <x,y,z> = 8 we know point (2,0,1) is on the plane so we have <2a1,4,0> dot (2,0,1)= 2(2a1) which means 2(2a1)= 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

phi
 one year ago
Best ResponseYou've already chosen the best response.3Find points on xyze^x+y^2=3 where.... I am getting (0,2,5/2) and (0,2,5/2)

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1isn't the dot product for two vectors? how can we do it with a vector and a dot

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, a dot product is between two vectors. But exactly what is the question? can you ask your question a different way?

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1I don't understand this part N dot P = d where P is any point on the plane <2a1,4,0> dot <0,2,a> = 8 thus d= 8 we know point (2,0,1) is on the plane so we have <2a1,4,0> dot (2,0,1)= 2(2a1) why is <2a1,4,0> dot (2,0,1) == <2a1,4,0> dot <0,2,a>

phi
 one year ago
Best ResponseYou've already chosen the best response.3I am using the "vector equation" for a plane perhaps you use \[ a (xx_0) + b(yy_0) +c (zz_0) = 0 \] ?

phi
 one year ago
Best ResponseYou've already chosen the best response.3if 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 \[ < a,b,c> \cdot <x,y,z> = d \]

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1Thank you!!! This makes perfect sense now. I can't thank you enough!

phi
 one year ago
Best ResponseYou've already chosen the best response.3and 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>

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1If 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.

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1But thank you! I've been strugling with this one the whole day.

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes, it was a bit painful. (especially when I wrote the d/dy of y^2 as 2y but with my handwriting, thought it was zy... that caused some grief)

Castiel
 one year ago
Best ResponseYou've already chosen the best response.1happened to me a bunch of times too. 2's and z's don't go well together.
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