Partial derivatives; Tangent Planes

JC3187
Messages
15
Reaction score
0
Hi guys,

Question is:

Find the slopes of the curves of intersection of surface z = f(x,y) with the planes perpendicular to the x-axis and y-axis respectively at the given point.

z = 2x2y ...at (1,1).

fx(x,y) = 4xy ∴ Slope = 4
fy(x,y) = 2x2 ∴ Slope = 2

Is this wrong?

Answer has it the other way around fx = 2 and fy =4

what mistake did i make?

Thanks
 
Physics news on Phys.org
JC3187 said:
Hi guys,

Question is:

Find the slopes of the curves of intersection of surface z = f(x,y) with the planes perpendicular to the x-axis and y-axis respectively at the given point.

z = 2x2y ...at (1,1).

fx(x,y) = 4xy ∴ Slope = 4
fy(x,y) = 2x2 ∴ Slope = 2

Is this wrong?

Answer has it the other way around fx = 2 and fy =4

what mistake did i make?

Thanks
Your work looks fine. It looks like they switched the values in the "answer" you showed.
 
  • Like
Likes 1 person
I thought so too, thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top