High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Tangent Planes, Gradient Vectors, and the Normal Line

Tangent Planes & Linear ApproximationPrevious Section
Maximum and Minimum Points - Finding Critical Points in Level CurvesNext Section
Popular Courses
Find My Course
0:00 / 0:00

Gradient Vectors, Tangent Planes, Normal Lines

Let us look at tangent planes in terms of the gradient vector and the normal line.

Wize Concept
~ The gradient vector ∇f(x0, y0) \nabla f(x_0,~y_0)~is orthogonal to the level curve f(x, y)=k, k∈ℜ, f(x,~y)=k,~k\in\Re,~at the point (x0, y0)(x_0,~y_0).

~ The gradient vector ∇f(x0, y0, z0) \nabla f(x_0,~y_0,~z_0)~is orthogonal to the level curve f(x, y, z)=k, k∈ℜ, f(x,~y,~z)=k,~k\in\Re,~at the point (x0, y0, z0)(x_0,~y_0,~z_0).

~ The gradient vector is given by ∇f(x, y, z)= <fx, fy, fz>.\nabla{f(x,~y,~z)}=~<f_x,~f_y,~f_z>.

So, the tangent plane to f(x, y, z)=k f(x,~y,~z)=k~is:
fx(x0, y0, z0)(x−x0)+fy(x0, y0, z0)(y−y0)+fz(x0, y0, z0)(z−z0)=0\boxed{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)=0}

Sometimes, we want a line orthogonal to a surface at a point called the normal line and it is parallel to the gradient vector at the point (x0, y0, z0). (x_0,~y_0,~z_0).~

Thus, the equation of the normal line is:

r(t)⃗=<x0,y0,z0>+t⋅}∇f(x0,y0,z0)\boxed{\vec{r(t)}=<x_0,y_0,z_0>+t\cdot\}\nabla{}f(x_0,y_0,z_0)}

Example

Find the tangent plane and the normal line to x216+y2100−z264=1 at (4, 5, −4)\frac{x^{2}}{16}+\frac{y^{2}}{100}-\frac{z^{2}}{64}=1~\text{at}~(4,~5,~-4).

Let F(x, y, z)=x216+y2100−z264. Then, F(x,~y,~z)=\frac{x^2}{16}+\frac{y^2}{100}-\frac{z^2}{64}.~\text{Then,}~
∇F= <x8, y50, −z32>∇F(4, 5, −4)=<12, 110, 18>\nabla F=~\Big<\frac{x}{8},~\frac{y}{50},~-\frac{z}{32}\Big>\newline{}\newline{} \nabla F(4,~5,~-4)=\Big<\frac{1}{2},~\frac{1}{10},~\frac{1}{8}\Big>
The tangent plane is:
12(x−4)+110(y−5)+18(z+4)=0\frac{1}{2}(x-4)+\frac{1}{10}(y-5)+\frac{1}{8}(z+4)=0
The normal line is:
r(t)⃗= <4, 5, −4>+t<12, 110, 18>\vec{r(t)}=~<4,~5,~-4>+t\Big<\frac{1}{2},~\frac{1}{10},~\frac{1}{8}\Big>

0:00 / 0:00

Example

Find tangent plane and normal line to x2y=4zex+y−35 at (1, −1, 1).x^2y=4ze^{x+y}-35~\text{at}~(1,~-1,~1).

Let f(x, y, z)=4zex+y−x2yf(x,~y,~z)=4ze^{x+y}-x^2y. We need to find fx(1, −1, 1), fy(1, −1, 1), and fz(1, −1, 1).f_x(1,~-1,~1),~f_y(1,~-1,~1),~\text{and}~f_z(1,~-1,~1).

fx=4zex+y−2xyfy=4zex+y−x2fz=4ex+yfx(1, −1,1)=6fy=3fz(1, −1, 1)=4∴ ∇f(1, −1,1)=<6, 3, 4>\begin{array}{c|c|c} f_x=4ze^{x+y}-2xy &f_y=4ze^{x+y}-x^2 & f_z=4e^{x+y} \\\\ f_x(1,~-1,1)=6 & f_y=3 & f_z(1,~-1,~1)=4 \end{array}\newline{}\newline{} \therefore~\nabla f(1,~-1,1)=<6,~3,~4>
Tangent Plane:
6(x−1)+3(y+1)+4(z−1)=06x+3y+4z=76(x-1)+3(y+1)+4(z-1)=0\newline 6x+3y+4z=7
Normal Line:
r(t)⃗= <1, −1, 1>+t<6, 3, 4>\vec{r(t)}=~<1,~-1,~1>+t<6,~3,~4>

Practice

An ant is crawling along a flat concrete surface with the temperature of the surface at a point (x, y) given by T=x3e4x2+y3T=x^3e^{4x^2+y^3}.

a) At (3, -1), what is the direction of the greatest decrease of the temperature?


b) Find the rate of change of the temperature if the ant crawled from (3, -1) at speed 1 in the direction of the vector <10, −4>.<10,~-4>.