Study guide and
10 practice problems
on:
Surfaces in 3d
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Study Guide
Tangent vectors to surfaces
(4 problems)
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)
Normal vectors to surfaces
(5 problems)
A normal vector to the surface $g(x,y,z)=0$ at $(x,y,z)$ is given by $\nabla g(x,y,z)$.
(5 problems)
A normal vector to the explicitly defined surface $z = f(x,y)$ is $$\left \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \right \rangle$$
(1 problem)
Tangent planes to surfaces
(2 problems)
Sketching a 3d surface
(1 problem)
Related topics
Sketching a 3d surface
(1 problem)
Tangent vectors to surfaces
(4 problems)
Normal vectors to surfaces
(5 problems)
Tangent planes to surfaces
(2 problems)
Multivariable calculus
(147 problems)
Functions of several variables
(36 problems)
Practice problems
Sketch the three-dimensional surface and level curves of $z = y^2 + x$.
Solution
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
Find a normal vector to the surface $x^3 + y^3 z = 3$ at the point $(1,1,2)$.
Solution
Find the tangent plane at $(1,1,1)$ to the surface $$x^2 + y^2 + z^2 + xy + xz = 5.$$
Solution
Give the equation for the tangent plane to the surface $z x^2 + x y^2 + y z^2 = 5$ at the point $(-1,1,2)$.
Solution
Use 3d level surfaces to show that a normal vector to $z=f(x,y)$ is given by $\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \rangle$.
Solution
Find a normal vector to the curve $\sqrt{x} + \sqrt{y} = 2$ at $(x,y) = (1,2)$. Use it to find the tangent line at $(1,1)$ expressed in the form $\bfn \cdot \bfx = b$.
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