Find the potential function f for the field F. F = 2xe^(x^2+y^2) i + 2ye^(x^2+y^2) j f(x, y, z) = e^(x^2+y^2) + C f(x, y, z) = 2e^(x^2+y^2) + C f(x, y, z) = (e^(x^2+y^2))/2 + C f(x, y, z) = e^(x^2) + e^(y^2) + C
Added by Ines D.
Close
Step 1
We want to find a potential function f(x, y) such that: ∇f(x, y) = F This means that: ∂f/∂x = 2xe^(x^2+y^2) ∂f/∂y = 2ye^(x^2+y^2) Now, we can integrate each partial derivative with respect to their respective variables: ∫(∂f/∂x) dx = ∫(2xe^(x^2+y^2)) Show more…
Show all steps
Your feedback will help us improve your experience
Madhur L and 65 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find a potential function $f$ for the field $\mathbf{F}$. $$\begin{aligned} &\mathbf{F}=\left(\ln x+\sec ^{2}(x+y)\right) \mathbf{i}+\\ &\left(\sec ^{2}(x+y)+\frac{y}{y^{2}+z^{2}}\right) \mathbf{j}+\frac{z}{y^{2}+z^{2}} \mathbf{k} \end{aligned}$$
Integrals and Vector Fields
Path Independence, Conservative Fields, and Potential Functions
Find a potential function $f$ for the field $\mathbf{F}$. $$\begin{aligned} &\mathbf{F}=\frac{y}{1+x^{2} y^{2}} \mathbf{i}+\left(\frac{x}{1+x^{2} y^{2}}+\frac{z}{\sqrt{1-y^{2} z^{2}}}\right) \mathbf{j}+\\ &\left(\frac{y}{\sqrt{1-y^{2} z^{2}}}+\frac{1}{z}\right) \mathbf{k} \end{aligned}$$
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD