Study guide and
4 practice problems
on:
Tangent vectors to surfaces
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A tangent vector to a surface is perpendicular to a normal vector to that surface.
A tangent vector to a surface has a slope (rise in $z$ over run in $xy$) equal to the directional derivative of the surface height $z(x,y).$
Study Guide
Using directional derivatives to find a tangent vector to a surface
(3 problems)
Using a normal vector to find a tangent vector to a surface
(1 problem)
Related topics
Using a normal vector to find a tangent vector to a surface
(1 problem)
Using directional derivatives to find a tangent vector to a surface
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Surfaces in 3d
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Multivariable calculus
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Functions of several variables
(36 problems)
Practice problems
Consider the surface $z = 10 - x^2 - 2 y^2$. At $(1,-1,7)$, find a 3d tangent vector that points in the direction of steepest ascent.
Solution
Consider the surface $z = f(x,y)$. What is the 3d tangent vector at $(x_0,y_0, f(x_0, y_0))$ that has $(x,y)$ components $a \mathbf{i} + b \mathbf{j}$?
Solution
Consider the surface $x^3 + y^3 z = 3$. Find tangent vector at the point $(1,1,2)$ that has $\mathbf{i}$ component 1 and $\mathbf{j}$ component 1. To find it, first find a normal vector.
Solution
Consider the surface given by $z = x^2 + y^2$. What is the (3d) tangent vector at $(1,1)$ that has an $\mathbf{i}$ component of 0 and a $\mathbf{j}$ component of 2?
Solution