find a normal vector at each point(x,y,z) for the cone z^2=x^2+Y^2 for 0

Question

find a normal vector at each point(x,y,z) for the cone z^2=x^2+Y^2 for 0<=z<=1

rational · Answer

normal vector  =  $\langle f_x, f_y, -1 \rangle$

rational · Answer

find the partials and plugin

anonymous · Answer

so my answer would be <2x,2y,-1>?

rational · Answer

$z^2 = x^2 + y^2  \implies f_x =    \dfrac{x}{z}, f_y = \dfrac{y}{z}$

then the normal vector wud be :  $\left< \dfrac{x}{z}, \dfrac{y}{z}, -1\right>$

anonymous · Answer

i derrived and im supposed to integrate right? or how did you get x/z

rational · Answer

$z^2 = x^2 + y^2$

 $f_x = \dfrac{\partial z}{\partial x} = ?$

 $f_y = \dfrac{\partial z}{\partial y} = ?$

anonymous · Answer

what does that mean sorry

anonymous · Answer

i have been up all night

rational · Answer

you familiar with taking partial derivatives, right ?

anonymous · Answer

yes very im just going blank right now

anonymous · Answer

fx means derrivae f in terms of x

rational · Answer

yep !

rational · Answer

$z^2 = x^2 + y^2$

partially differentiate both sides in terms of "x"

rational · Answer

what do u get ?

rational · Answer

$z^2 = x^2 + y^2$

partially differentiate both sides in terms of "x" :

$2z \dfrac{\partial z}{\partial x} =  2x + 0$

rational · Answer

right ?

anonymous · Answer

why do you have 2z i get the 2x

anonymous · Answer

is it the (df/dx)(dx/dz) thing?

rational · Answer

no lol, we're just differentiating implicitly

rational · Answer

lets try it directly otherwise

rational · Answer

$z^2 = x^2 + y^2$


$z =  \sqrt{x^2 + y^2}$

rational · Answer

Now find your partials  $f_x$ and $f_y$

anonymous · Answer

wtf i have no idea whats going on

rational · Answer

your function is :

$z = f(x,y) =  \sqrt{x^2+y^2}$

$f_x = ?$
$f_y = ?$

anonymous · Answer

so f(x,Y)=x+y
then

anonymous · Answer

hold on

anonymous · Answer

for fx i got$$x \div \sqrt{x ^{2+y ^{2}}}$$

anonymous · Answer

since z=sqrt(x^2+y^2)

anonymous · Answer

you get x/z

rational · Answer

do u mean :

$f_x =   \dfrac{x}{\sqrt{x^2+y^2}} =  \dfrac{x}{z}$ ?

anonymous · Answer

yeah

rational · Answer

Yep ! you got it !!

anonymous · Answer

lol i had a big brain fart

anonymous · Answer

can you help me with something more complicated?

rational · Answer

il try, ask..

anonymous · Answer

calculate the surface area of the part of the paraboloid z=6-x^2-y^2 above the plane with equation z=2

rational · Answer

straightforward problem

rational · Answer

surface area :

$$\iint \limits_R   \sqrt{f_x^2 + f_y^2 + 1} ~dA$$

anonymous · Answer

z=6-r^2

anonymous · Answer

oh i know that lol but my limits of integration

rational · Answer

for gettign the limits, sketch the region quick

anonymous · Answer

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