## 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

1. Castiel

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

2. 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

3. Castiel

@Vocaloid can you pls help

5. Vocaloid

it's been a while since I've done a problem like this @phi @ganeshi8

@phi

@ganeshie8

@LynFran

9. 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

10. phi

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

11. ganeshie8

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

12. phi

yes, that was the wrong thought

13. Castiel

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

14. ganeshie8

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

15. Castiel

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

16. ganeshie8

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

17. ganeshie8

Yes..

18. 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

19. 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)

20. 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

21. 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>

@phi thank you so much for helping

24. 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 <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

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

26. Castiel

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

27. phi

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

28. 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>

29. phi

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

30. Castiel

Ohhh yeah I see

31. 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 $< a,b,c> \cdot <x,y,z> = d$

32. Castiel

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

33. 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 <x,y,z>

34. 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.

35. Castiel

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

36. 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)

37. Castiel

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