Second example of line integral of conservative vector field | Multivariable Calculus | Khan Academy

TL;DR

Calculate a line integral over a curve that is not closed using vector valued functions.

Transcript

Let's do another problem. Very similar to the last one, but with a subtle difference. And that subtle difference will make a big difference. Let's say we take the line integral over some curve c-- I'll define the curve in a second-- of x squared plus y squared dx plus 2xy dy-- and this might look very familiar. This was very similar to what we saw ... Read More

Key Insights

• 😚 The line integral over a curve that is not closed requires a different approach than a closed line integral.
• 🏑 To determine if a vector field is conservative, we need to find a potential function whose gradient equals the vector field.
• ❓ The potential function can be found by solving the partial differential equations for the partial derivatives of the potential function.
• 🫥 Evaluating the potential function at the endpoints of the curve allows us to calculate the line integral.
• 🫥 If the vector field is conservative, the line integral is path independent and only depends on the endpoints of the curve.

Questions & Answers

Q: What is the difference between this line integral problem and the previous one?

The previous problem involved a closed line integral, while this problem involves a curve that is not closed.

Q: Can we apply the same method as before to determine if the vector field is conservative?

No, since the curve is not closed, we cannot use the fact that a closed loop integral is equal to zero.

Q: How can we determine if the vector field is conservative in this case?

By finding the potential function f(x,y) such that its gradient (the vector field) is equal to f(x,y).

Q: How can we find the potential function for this vector field?

To find the potential function, we solve the partial differential equations ∂f/∂x = x^2 + y^2 and ∂f/∂y = 2xy.

Summary & Key Takeaways

• The problem involves calculating a line integral over a curve using the vector field f(x,y) = x^2 + y^2 dx + 2xy dy.

• The curve, parameterized as x = cosine(t) and y = sine(t), starts at t = 0 and goes to t = pi.

• The goal is to determine if the vector field is conservative and find the potential function f(x,y).

• By evaluating the potential function at the endpoints of the curve, the line integral can be calculated.