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So the question is: Give the equation of the tangent plane at the surface level of the function $$f(x, y, z)=\cos (x+2 y+3 z)$$ at the point $(x, y, z)=(\pi / 2, \pi, \pi)$

I don't actually want you to calculate it (but if you like, the bettter :) ), I really want to know if my thought process is correct (If I didn't make a mistake e.g. my equation for a tangent plane).

So first I calculated the partial derivatives (at the given point)

$f_x=1$ $f_x=2$ $f_x=3$

and then I used the equation for a tangent plane (don't really know if this is correct) $$z=f(a,b,c)+f_x(a,b)(x-a)+f_y(a,b)(y-a)+f_z(a,b)(z-a)$$ and got $x+2y+3z-\frac{11}{2}\pi$

If you can help me, it would be very much appreciated.

Thanks in advance

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  • $\begingroup$ $$x+2y+3z=\frac{11}{2}\pi$$ $\endgroup$
    – Raffaele
    Dec 29, 2020 at 18:46
  • $\begingroup$ If you're going to find the tangent plane of the graph $w=f(x,y,z)$ (which is the formula you're trying to use here), you need another variable. Try not to confuse the graph of a function with a level set of the function. $\endgroup$ Dec 29, 2020 at 19:40

2 Answers 2

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This is asking you to first find the level surface of the function $f(x,y,z)$ that contains the point $(\pi/2,\pi,\pi)$. This would just be the surface implicitly defined by $0=\cos(x+2y+3z)$. The correct equation for the tangent plane of an implicitly defined surface is $$0=f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c).$$

The equation you have in the question is a combination of this equation and the tangent plane of an explicitly defined surface $z=f(x,y)$, which is $$z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

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  • $\begingroup$ Thank you so much $\endgroup$
    – TheCreator
    Dec 29, 2020 at 19:24
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This is just alternative solution without using Calculus

Notice that $f(x,y,z) = k_{1} \implies x+2y+3z=k_{2}$ i.e. the tangent plane equation is simply $x+2y+3z=\frac{\pi}{2}+2\pi+3\pi$

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