449014…: “Nearer the noisy sphere”

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Suppose that $f(x,y) = \sin (2x+4y)$ . Find the equation of the tangent plane to the surface $z=f(x,y)$ at the point where $(x,y)=(2,-1)$ . Express your final answer in the form $Ax+By+Cz=D$ .

$\answer {2} \cdot x + \answer {4} \cdot y - z =\answer {0}$

The point on the plane can be found by evaluating $f(x,y)$ at $(2,-1)$ . Doing so, we find $f(2,-1) = \answer {0}$ .
To construct a normal vector, note that
- $\pp [f]{x}(x,y) = \answer {2\cos (2x+y)}$ so $\pp [f]{x}(2,-1) = \answer {2}$ .
- $\pp [f]{y}(x,y) = \answer {4\cos (2x+y)}$ so $\pp [f]{y}(2,-1) = \answer {4}$ .
Thus, a normal vector to the tangent plane is $\vec {n} = \vector {\answer {2},\answer {4},-1}$ .

We now compute the tangent plane. Recall that the equation of the plane with normal vector $\vec {n}=\vector {a,b,c}$ that passes through $P_0 = \big (x_0,y_0,z_0\big )$ is

$a(x-x_0)+b(y-y_0)+c(z-z_0) = 0.$