Question

will award medal
Find an equation of the tangent plane to the given surface at the specified point.
z = sqrt(xy)

Answer 1

p=(3,3,3)

Answer 2

@mathmate

Answer 3

so i keep getting the answer of z=2x+2y-9, from the partial of y and the partial of x evaluated at 3,3,3. and using the fact that z-z_0=f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)

Answer 4

@SithsAndGiggles

Answer 5

but my online says its worng

Answer 6

You have
\[f(x,y)=\sqrt{xy}~~\implies~~\nabla f(x,y)=\left(\frac{\sqrt y}{2\sqrt x},\,\frac{\sqrt x}{2\sqrt y}\right)\]
So at \((x,y)=(3,3)\), you get \(\nabla f(3,3)=\left(\dfrac{1}{2},\dfrac{1}{2}\right)\).

The tangent plane would then be
\[z-3=\frac{1}{2}(x-3)+\frac{1}{2}(y-3)\]

Answer 7

sorry that coding is really confusing

Answer 8

Is the latex not showing up?

Answer 9

no not for me

Answer 10

oh....

Answer 11

do its not regular differentiating partially

Answer 12

What do you mean?

Answer 13

im just so confused on the partial derivitive. my teacher must have not clarified what it means

Answer 14

oh pellet so would you use the product rule and seperate

Answer 15

@mathmate
l

Answer 16

In case you were not clear about:
\(\nabla f(x,y)\)
It means
\(\nabla f(x,y)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial x}\right )=\left(\dfrac{\sqrt y}{2\sqrt x},\,\dfrac{\sqrt x}{2\sqrt y}\right)\)