Question

Calc III Tutorial: Tangent Planes & Differentials

Answer 1

\({\bf{Definitions:}}\)
tangent plane: the plane through P0 normal to \(∇f|_{P_{0}}\)
and is given by \[f_{x}(P_{0})(x-x_{0}) + f_{y}(P_{0})(y-y_{0}) + f_{z}(P_{0})(z-z_{0}) = 0\]
normal line: the line through P0 parallel to \(∇f|_{P_{0}}\)
and is given by the parametric equations
\[x = x_{0} + f_{x}(P_0)t\] \[y = y_{0} + f_{y}(P_0)t\] \[z = z_{0} + f_{z}(P_0)t\]

Answer 2

basic procedure: find the gradient of the function, evaluate at the given point, plug the gradients and (x0,y0, etc.) values into the tangent plane equation and normal plane equation as desired

Answer 3

at the surface z = f(x,y) the tangent plane becomes

\[f_{x}(x0,y0)(x-x0) + f_{y}(x0,y0)(y-y0) + (z-z0) = 0\]
this equation is derived by letting f(x,y) - z = 0 and applying the tangent plane formula

special case: at the intersection of two surfaces, the tangent line at the intersected surface can be found by taking the gradient of both surfaces, finding the cross product, and creating a parametric equation using the point of intersection and the newly generated cross product

Answer 4

\({\bf{Estimating~Directional~Change:}}\)

change in a function wrt to a particular direction can be found by multiplying the directional derivative with the infinitely small distance ds. recall that the directional derivative is the gradient at a point, times the unit vector of the direction vector.

\[df = (∇f|_{P_{0}}·u)ds\]

\({\bf{Linearization:}}\) allows us to estimate a nonlinear function as a tangent-line
in multivariable calculus this is given as:

\[L(x,y) = f(x0,y0) + f_{x}(x0,y0) + f_{y}(x0,y0)\]
which will give out a linear function that approximates the original across a limited domain around a point

the error of this approximation is given as

|E(x,y)| <= (1/2)M(|x-x0| + |y-y0|)^2 where M is the upper bound of the second partial derivatives fxx, fyy, and fxy

Answer 5

\({\bf{Differentials:}}\) as a reminder, the differential of a function is the change in a function across some difference in domain values
in multivariable calculus this is given as

\[df = f_{x}(x0,y0)dx + f_{y}(x0,y0)dy\]

using a more specific example, Volume gives us

\[dV = V_{r}(r0,h0)dr + V_{h}(r0,h0)dh\]
in this particular example you would typically be given an initial radius, initial height, and given dr and dh so you would really only have to calculate the partial derivatives at r0,h0

Answer 6

Source material is section 14.6 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.