Question

Find the point on the paraboloid z=9-(X^2)-(y^2) at which the tangent plane is parallel to the plane z=4y Please!

Answer 1

unless I am mistaken we need the gradient vectors of each surface
when those vectors are parallel, so are the planes

Answer 2

The problem is that I dont have (or rather KNOW) the points to use when finding the gradient vectors.

Answer 3

you don't need a point to find the gradient; you can get a formula in terms of variables:
rewrite the parabaloid\[f(x,y,z)=x^2+y^2+z=9\]\[\nabla f=<2x,2y,1>\]now do the same for the plane\[g(x,y,z)=-4y+z=0\]\[\nabla g=<0,-4,1>\]these vectors are parallel at\[x=0, y=-2,\text{ and }z=t;t\in\mathbb R\]hm... looks like they are parallel at infinite points unless I made a mistake

Answer 4

I also thought of the same answer but couldnt write it down. I think that is correct.Thanks

Answer 5

welcome :D

Answer 6

Can I send my question to a specific person direclty here?

Answer 7

you can tag a person
@Peterson0000 for example
and they will receive a message

Answer 8

Thanks. :D looks like I'll be tagging you a lot. do you maybe have links to good online material I can study calculus from.

Answer 9

sure thing, I am a big fan of Paul's online math notes:
http://tutorial.math.lamar.edu/
here is the calcIII section:
http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx
though it is not as thorough as MIT's open course ware
http://ocw.mit.edu/index.htm
however OCW is more time-consuming

Answer 10

Thanks. I got another question can I ask?
Sorry to be so bothering....

Answer 11

no bother, it's why I'm here :D

Answer 12

Find the point on the surface z=2(x^2)-(y^2)+3y where the normal is parallel to theline that joins the point A(1,1,2) and B(9,-4,3)

Answer 13

we need the vector that points from A to B (or vice-versa)
can you find that?

Answer 14

AB<8,-5,1>

Answer 15

right, now we need the gradient of the surface
see what you get for that

Answer 16

If you are having trouble please specify where

Answer 17

Gradient vector f = < 4x,(-2y+3),1>

Answer 18

there is a problem with your signs
before you find the gradient of f you need f in the form\[f(x,y,z)=k\]where k is a constant
I think you just took the gradient with variables on both sides of the = sign, which does not work
so rewrite this as f(x,y,z)=k, then take the gradient again

Answer 19

< 4x, (-2y+3),-1>

Answer 20

ok, that looks right
now we need to find when the components of each vector are equal (because that will be when they are parallel)
so we need when 8=4x, -5=-2y+3, 1=-1
but hm... 1=-1 is not possible, so what do we do?

Answer 21

here we need to remember that making each component of a vector negative will not change it's orientation in space, only the way (forwards or backwards) that it points
therefor it will be parallel to the same vectors either way
hence, we can make all the components of the gradient negative and we will be able to find an answer

Answer 22

also note that you get the same result by moving everything over to the other side of the equals sign before taking the gradient, which is a perfectly valid move of course
f(x,y,z)=-2x^2+y^2-3y+z
grad f=(-4x,2y-3,1)
now we can set the components of this and the other vector equal

Answer 23

\[-4x=8\to x=-2\]\[2y-3=-5\to y=-1\]\[1=1\to z=t;t\in\mathbb R\]so the points are\[(-2,-1,t);t\in\mathbb R\]which is a line along the z-axis
I now just realized that I think we never finished our other problem...

Answer 24

in order to get the z value we need to find for which value of z this is actually on the surface
hence we need to plug these numbers into f to solve for z

Answer 25

\[f(x,y,z)=-2x^2+y^2-3y+z=0\]\[f(-2,-1,z)=-8+1+3+z=0\to z=4\]hence the point is\[(-2,-1,4)\]you think you can go back and finish the first problem with that info?

Answer 26

z=5 therefor points are (0,-2,5) Thanks a lot

Answer 27

right, awesome!
glad I could help :D

Answer 28

enjoy your evening I have to go now. Thanks again. I learnt a lot tonight