Question

what is d/dx of sqr root (1−x2−y2)?

Answer 1

Rupesh get this it is from microsoft and will solve everything for you
http://www.microsoft.com/download/en/details.aspx?id=15702

Answer 2

\[\frac{-x}{\sqrt{-x^2-y^2+1}}\]

Answer 3

= -(-2x)/ 2sqrt(1−x2−y2) = x/sqrt(1−x2−y2)

Answer 4

Better: buy a ti-89 and it will even do linear algebra for you!

Answer 5

agadgd that program will do linear too

Answer 6

\[d/dx \sqrt{1−x^2−y^2}\]
\[\frac{d/dx(1−x^2−y^2)}{\sqrt{1−x^2−y^2}}\]
\[\frac{−2x−2y\ y'}{\sqrt{1−x^2−y^2}}\]

Answer 7

forgot the 2 on the bottom :)

Answer 8

\[\frac{−2x−2y\ y'}{2\sqrt{1−x^2−y^2}}\]
\[\frac{2(−x−y\ y')}{2\sqrt{1−x^2−y^2}}\]
\[\frac{−x−y\ y'}{\sqrt{1−x^2−y^2}}\]
and solve for y'

Answer 9

But y′ is zero, since we are differentiating with respect to x. Isnt it?

Answer 10

no; y' is unknown

Answer 11

but to clarify; is this partial derivative or implicit?

Answer 12

Implicit

Answer 13

then y' pops out; dy/dx doesnt = 0 all on its own

Answer 14

its simply a rate of change in y with respect to x

Answer 15

ok. I understand

Answer 16

But my doubt started when I was trying to find the outward normal vector to a sphere. Can you help me with it?

Answer 17

\[\frac{d}{dx} \sqrt{1−x^2−[y(x)]^2}\]
\[\frac{{d}{dx}(1−x^2−[y(x)]^2)}{2\sqrt{1−x^2−[y(x)]^2}}\]
\[\frac{\frac{d}{dx}1−\frac{d}{dx}x^2−\frac{d}{dx}[y(x)]^2}{2\sqrt{1−x^2−[y(x)]^2}}\]
\[\frac{−2x−2[y(x)]*\frac{dy}{dx}}{2\sqrt{1−x^2−[y(x)]^2}}\]
\[\frac{2(−x−[y(x)]*\frac{dy}{dx})}{2\sqrt{1−x^2−[y(x)]^2}}\]
\[\frac{−x−[y(x)]*\frac{dy}{dx}}{\sqrt{1−x^2−[y(x)]^2}}\]
\[\frac{−x−y*\frac{dy}{dx}}{\sqrt{1−x^2−y^2}}\]

Answer 18

outward normal vector to a sphere is the partial of df/dx and df/dy

Answer 19

agreed. But in my previous question on that , somebody replied that to be the grad of that function.

Answer 20

But grad of that function and the the partial of df/dx and df/dy are turning out to be different. So I am confused?

Answer 21

the gradient is the "equation" for the normal; same thing

Answer 22

Sorry for taking too much of your time. In the question it was to find the direction of the outward normal to x2 + y2 + z2=1 at (1,1,1)? Kindly refer my previous question. Somebody replied me to take the grad of the function, which got to be 2 x r (where r is the radius vector)
But when we take in parametric form by taking partial differentiation, are we getting the same answer. Confused. kindly help..

Answer 23

\[f(x,y)= \sqrt{1−x^2−y^2}\]
\[df/dx= \frac{-x}{\sqrt{1−x^2−y^2}}\]
\[df/dy= \frac{-y}{\sqrt{1−x^2−y^2}}\]

partials are done by treating the non-variable a constant

the normal is: <df/fx, df/dy> for any given (x,y)

Answer 24

internet is goofing up on my end

Answer 25

if we take the partials of the f(x,y,z) the are simliar and the same concepts are involved

Answer 26

the gradient provides general equations that are used to solve for the normal at any given point; so if we know the gradient, and a given point; we can know the normal at the point ...

Answer 27

Oh. Ok.

Answer 28

"In the question it was to find the direction of the outward normal to x2 + y2 + z2=1 at (1,1,1)"

\[f(x,y,z) = x^2 + y^2 + z^2 -1\]
\[df/dx = 2x\]
\[df/dx = 2y\]
\[df/dx = 2z\]

normal at (1,1,1) when the gradient is: 2<x,y,z> = 2<1,1,1>

Answer 29

Okay. Much better understanding now.

Answer 30

it just so happens that the vector from the origin (0,0,0) to the point (1,1,1) is: <1,1,1>

and in sphere, the normal is in a straight line with the vector from the origin :) if that makes sense

Answer 31

Okay. Much better understanding now.

Answer 32

Thanks very much. Highly appreciated.

Answer 33

youre welcome :)