12.7.17 pg. 714

For a function z=f(x,y), a point P is given. Find the equation of the tangent plane to f at P.

  1. \(f(x,y)=2x^2y-4xy^2\), \(v=<1,3>\) and \(P=(2,3)\)
# Plot of f(x,y)
x = y = seq(-10, 10, length= 30)
f = function(x,y){z = (2*(x^2)*y)-(4*x*y^2)}
z = outer(x,y,f)
persp(x, y, z, theta = 30, phi = 30, expand = 0.5, col = "lightblue")

The plane through P with a normal vector n is the tangent plane to f at P. It’s standard form is \(a(x-x_0)+b(y-y_0)-(z-f(x_0,y_0))=0\)

\(f(x_0,y_0))=f(2,3)=2(2^2)(3)-4(2)(3^2)=24-72=-48\)

with Point P, the standard form becomes \(a(x-2)+b(y-3)-(z-(-48))\)

And now we find the constants a and b. To do that, we have to find the partial derivatives of the function evaluated at the point P.

\(a=\frac{df}{dx}(2,3)=\frac{df}{dx}(2x^2y)-\frac{df}{dx}(4xy^2)=4yx-4y^2\) at P(2,3) \(a=4(3)(2)-4(3^2)=-12\)

\(b=\frac{df}{dy}(2,3)=\frac{df}{dx}(2x^2y)-\frac{df}{dx}(4xy^2)=2x^2-8xy\) at P(2,3) \(a=2(2^2)-8(2)(3)=-40\)

Making the equation of the tangent plane: \(-12(x-2)-40(y-3)-(z+48)=0\)

Valuable elements from this course:

I think the majority of the class can agree that this was a good refresher (and even introduction to some concepts that might not have been addressed to some based on different backgrounds). What really was valuable was actually using the concepts in practice (especially the matrices). If given the time, more attention to the integration of theory and real-life problems would have been great (some of Nathaniel’s discussions and posts were so awesome because they did just that). And, finally, I’m going to go ahead and say that I liked the discussions. I don’t know if they count as an “element from this course”, but they definitely helped me learn by going through others’ problems or getting help on mine or even just seeing what someone thought was interesting.