Fields that only care about endpoints

Mathematics · Calculus & Analysis · ages 20-21

Name ______________________ Date ____________

P and Q have continuous first partials on an open region where every loop contracts to a point inside the region. Which check tests whether F = (P, Q) is conservative?
- Check whether dP/dx = dQ/dy throughout the region
- Check whether dQ/dx = dP/dy throughout the region
- Check whether dP/dx = -dQ/dy throughout the region
A field has continuous first partials and integral zero around one closed loop. This alone proves that it is conservative throughout its open, connected domain.
Circle one: True False
Two potentials for the same field are f and h = f + 8. What is [h(B) - h(A)] - [f(B) - f(A)]?
Answer: ______________
Which property defines a conservative vector field on its domain?
- It equals grad f for a function f throughout its domain
- It has equal components at all points within its domain
- It has equal arrow lengths at every point of its domain
Let f(x, y) = x²y + y. Find the integral of grad f along any piecewise smooth curve from (0, 0) to (1, 1).
Answer: ______________
For F = (2xy, x² + 1) on the whole plane, integrating P in x gives f = x²y + g(y). Which step determines g'(y)?
- Match the y derivative of f with the component Q
- Match the x derivative of f with the component Q
- Match the y derivative of f with the component P
On the whole xy-plane, F = (4x + 3y, 3x + 6y). Which function is a potential for F?
- 2x² + 6xy + 3y²
- 4x² + 3xy + 6y²
- 2x² + 3xy + 3y²
- 2x² + 3xy - 3y²
Is F = (2xy, x² + 1) conservative on the whole xy-plane?
- No. P_y = 2xy and Q_x = 2x everywhere on the plane
- No. P_y = 2x and Q_x = x everywhere on the plane
- Yes. P_y = 2x and Q_x = 2x everywhere on the plane
On the whole plane, F = (2xy, x² + 2x). Sam claims f = x²y is a potential after checking only f_x = 2xy. What does checking f_y show?
- Matching f_x to P is enough to verify the entire gradient
- An added constant supplies the missing 2x in the y derivative
- f_y = x² differs from Q, so the candidate is not a potential
An object of fixed mass travels by two routes from A to B in a fixed Newtonian gravity field. U is its gravitational potential energy. Compare the work done by gravity, excluding other forces.
- Larger on the longer route because it covers more ground
- Equal on both routes: the work done by gravity is U(A) - U(B)
- Larger on the steeper route, since gravity opposes a steeper climb

Answer key

For grown-ups. Fold this page away before handing over the rest.

Fields that only care about endpoints W1-mt_5djKpI5F33-s1

Check whether dQ/dx = dP/dy throughout the region · Under the stated smoothness and domain assumptions, equality dQ/dx = dP/dy everywhere is the conservative-field test.
False · The characterization requires zero integral around every allowed closed loop. One selected loop does not establish it.
0 · Adding the same constant at both endpoints contributes 8 - 8 = 0, so the two endpoint differences are equal.
It equals grad f for a function f throughout its domain · A conservative field has a potential f whose gradient equals the field throughout its domain.
2 · f(1, 1) = 2 and f(0, 0) = 0. Subtract start from end: 2 - 0 = 2.
Match the y derivative of f with the component Q · The y derivative is x² + g'(y). Matching Q = x² + 1 gives g'(y) = 1.
2x² + 3xy + 3y² · For f = 2x² + 3xy + 3y², the derivatives f_x = 4x + 3y and f_y = 3x + 6y reproduce both components.
Yes. P_y = 2x and Q_x = 2x everywhere on the plane · P_y = Q_x = 2x everywhere. The polynomial components have continuous first partials, and every loop in the plane contracts within the plane.
f_y = x² differs from Q, so the candidate is not a potential · The candidate has f_y = x², but Q = x² + 2x. These are different functions, so the candidate does not reproduce F.
Equal on both routes: the work done by gravity is U(A) - U(B) · Gravitational work is path independent in this fixed field. Since F = -grad U, the work on either route is U(A) - U(B).