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

- Castiel

- chestercat

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- Castiel

Ok i solved for gradient \[grad=(y z-e ^{x})i+(xz+2y)j+(xz)k\]

- Castiel

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

- Castiel

My bad, the grad is actually grad=(yz−ex)i+(xz+2y)j+(xy)k, the last part is not xz but xy

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## More answers

- madhu.mukherjee.946

@Vocaloid can you pls help

- madhu.mukherjee.946

- madhu.mukherjee.946

- madhu.mukherjee.946

- Castiel

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

- phi

plug in (2,0,1) into the gradient

- ganeshie8

(2, 0, 1) need not exist on the given surface right

- phi

yes, that was the wrong thought

- Castiel

no it does not, when I plug 2,0,1 in grad I get (-e^2)i+2j

- ganeshie8

so the tangent plane looks like
\[(y_0z_0−e^{x_0})x+(x_0z_0+2y_0)y = d\]

- Castiel

yes, and d would equal to 2(y0z0-e^x0)

- ganeshie8

so the tangent plane looks like
\[(y_0z_0−e^{x_0})(x-2)+(x_0z_0+2y_0)y = 0\]

- ganeshie8

Yes..

- phi

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

- Castiel

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)

- phi

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

- phi

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>

- madhu.mukherjee.946

@Castiel is your problem solved

- madhu.mukherjee.946

@phi thank you so much for helping

- phi

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 = 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

- phi

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

- Castiel

isn't the dot product for two vectors? how can we do it with a vector and a dot

- phi

yes, a dot product is between two vectors. But exactly what is the question?
can you ask your question a different way?

- Castiel

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>

- phi

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

- Castiel

Ohhh yeah I see

- phi

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
\[ < a,b,c> \cdot = d \]

- Castiel

Thank you!!! This makes perfect sense now. I can't thank you enough!

- phi

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

- Castiel

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.

- Castiel

But thank you! I've been strugling with this one the whole day.

- phi

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)

- Castiel

happened to me a bunch of times too. 2's and z's don't go well together.

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