Study guide and
13 practice problems
on:
Level curves and surfaces
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The level curves of $f(x,y)$ are curves in the $xy$-plane along which $f$ has a constant value.
The level surfaces of $f(x,y,z)$ are surfaces in $xyz$-space on which $f$ has a constant value.
Sometimes, level curves or surfaces are referred to as level sets.
Related topics
Functions of several variables
(36 problems)
Multivariable calculus
(147 problems)
Practice problems
Sketch the level curves of $f(x,y) = xy$.
Solution
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
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
Let $f(x,y) = xy$.
a) Sketch the level curves of $f$.
b) Sketch the path of steepest descent starting at $(1,2)$.
b) Find the path of steepest descent starting at $(1,2)$.
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
Let $f(x,y) = x^2 + y^2$.
Describe the shape of the $f(x,y)=2$ level curve.
Without calculation, find the directional derivative at $(1,1)$ in the direction $-\bfi+\bfj$.
Hint: consider the level curve at $(1,1).$
By computation, find the directional derivative at $(1,1)$ in the direction of $-\bfi + \bfj$.
Solution
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