Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@Vocaloid can you pls help

Answer 5

it's been a while since I've done a problem like this

@phi @ganeshi8

Answer 6

@phi

Answer 7

@ganeshie8

Answer 8

@LynFran

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

yes, that was the wrong thought

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

Yes..

Answer 18

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

Answer 19

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)

Answer 20

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

Answer 21

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>

Answer 22

@Castiel is your problem solved

Answer 23

@phi thank you so much for helping

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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>

Answer 29

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

Answer 30

Ohhh yeah I see

Answer 31

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

Answer 32

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

Answer 33

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>

Answer 34

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.

Answer 35

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

Answer 36

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)

Answer 37

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