Equation for the tangent plane of z = sinx+cosy at (0,0)

Question

abb0t · Answer

Do you know the general formula? I believe there is one.

anonymous · Answer

fx (x-x0) + fy (y-y0) or something along those lines
I ended up getting cosx =0 which just looks wrong to me

abb0t · Answer

$z-f(x_0,x_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,x_0)(y-y_0)$

abb0t · Answer

cos(0) = 1

kc_kennylau · Answer

$$\begin{array}{rl}f(x,y)&=&\sin x+\cos y\f_x(x,y)&=&\cos x\f_y(x,y)&=&-\sin y\end{array}$$
$$\begin{array}{rl}\large f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0)(y−y_0)−(z−z_0)&=&0\\cos0\cdot x-\sin 0\cdot y-[z-(\sin0-\cos0)]&=&0\x-(z+1)&=&0\x-z-1&=&0\end{array}$$

kc_kennylau · Answer

P.S. It's my first day doing this so I'm not sure if I'm right or not. (actually I just searched about this in a few moments ago)

anonymous · Answer

Let me check over

abb0t · Answer

It's pretty straight forward, just find the partial with respect to x, and y. Then plug in the given conditions and you get an eqaution :)

kc_kennylau · Answer

sorry it should be $\sin0+\cos0$ which gives, at the end, $x-z+1=0$

anonymous · Answer

right. I think I've got it now. I'd give the medal to both of you if I could

anonymous · Answer

thank you both

kc_kennylau · Answer

I've given @abb0t a medal on behalf of you @rymdenbarn :)

anonymous · Answer

Thanks :p

kc_kennylau · Answer

no problem xD