partial derivatives: Find an equation for the plane

**break** · November 7th 2010, 04:49 AM

Find an equation for the plane tangent to the graph of z = 6 cos(xy) at the point (2, π/6, 3).

i went on to find
partial derivative x = -6 y sin(xy)
partial derivative y = -6 x sin(xy)

then i used the equation for tangent plane
$z=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)$

$z = 6cos(xy) + (-6)ysin(xy)(x-2)+(-6)xsin(xy)(y-\pi/6)$

$z = 6(.5) + (-6)(\frac{\pi}{6})sin(2 * \frac{\pi}{6})(x - 2) + (-6)(2)sin(2 * \frac{\pi}{6})(y - \frac{\pi}{6})$

$z = 3 + (-\pi)*\frac{\sqrt{3}}{2} * (x-2) + (-12)\frac{\sqrt{3}}{2}(y - \frac{\pi}{6})$

The thing is, i have no idea if i did this problem correctly or not. I haven't been taught how to deal with problems like this and trying to learn on my own. Can anyone help? Thanks!

**HallsofIvy** · November 7th 2010, 05:06 AM

You should learn how to check it yourself. If [tex]x= 2[/math and $y= \pi/6$ in your formula for the plane give z= 3? Is the "x-slope", the coefficient of x in the plane formula, equal to $-6(3) sin(2(\pi/6))$ ? Is the coefficient of y in theplane formula equal to $-6(\pi/6)sin(2(\pi/6))$ ?

Here, by the way is what I consider to be a simpler way to find the tangent plane to $z= f(x,y)$ . The function $g(x, y, z)= z- f(x,y)$ has $z- f(x,y)= 0$ or $z= f(x, y)$ as a "level surface". The gradient of g, $\nabla g$ is perpendicular to that level surface and so perpendicular to its tangent plane, the tangent plane to z= f(x,y).

If a normal to a plane at $(x_0, y_0, z_0)$ is [tex]\nabla g= <-f_x, -f_y, 1>[tex], then the plane is given by $-f_x(x_0, y_0)(x- x_0)- f_y(x_0, y_0)(y- y_0)+ (z- z_0)= 0$

**break** · November 7th 2010, 05:38 AM

so does that mean

$-6ysin(xy)(x-2)-6xsin(xy)(y-\frac{\pi}{6})-(z-3) = 0$
$-6(\frac{\pi}{6})sin(2*\frac{\pi}{6})(x-2)-6(2)sin(2*\frac{\pi}{6})(y-\frac{\pi}{6})-(z-3) = 0$
$-\pi*sin(\frac{\pi}{3})(x-2)-12sin(\frac{\pi}{3})(y-\frac{\pi}{6})-(z-3) = 0$

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