I'm lost on using the gradient for tangent planes.

answered 2014-07-10 07:25:20 -0600

Anonymous
740 ●2 ●27 ●49

updated 2014-07-10 09:38:26 -0600

I'm not completely sure what you're asking for but I think it is for an equation for the plane tangent to an $f(x)$. In section 14.5, question #11, it asks: Find an equation for the plane tangent to
$x^2-3y^2+z^2=7$ at $(1,1,3).$
In class, the professor provided us with the equation $$L(x)=f(x_0,y_0,z_0)+f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)$$. This is equation can also be expressed as $$f(x_0,y_0,z_0)=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)$$, which is the equation of a plane for a multivariate function. So using this equation, we are given $(x_0,y_0)$ as (1,1,3) and all we have to do is find the partial derivatives at this point and plug the information into the equation. It is almost time for class so I will post the rest of the procedure later this morning. Okay here is the step by step: find the partial derivatives $$f_x(x,y,z)=2x$$ $$f_y(x,y,z)=-6y$$ $$f_z(x,y,z)=2z$$ find the values at (1,1,3) $$f_x(1,1,3)=2$$ $$f_y(1,1,3)=-6$$ $$f_z(1,1,3)=6$$ plug the values into our given equation $$2(x-1)-6(y-1)+6(z-3)=0$$ since we have to take the partial derivatives of each variable, we can see that this is the gradient of the function since $$\bigtriangledown f(x,y,z)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$$