Work done by F = <2x, y> along y = 2x^2 from (0, 0) to (1, 2) Step by Step

Step by step

The question

Question: find the work done by F = ⟨2x, y⟩ along y = 2x² from (0, 0) to (1, 2).

F = ⟨2x, y⟩

along y = 2x² from (0, 0) to (1, 2)

1 Write the path in t

On the curve, let x be t.

Then y is 2t².

t goes from 0 to 1.

x = t, y = 2t²

So x = t

x = t

and y = 2t², with t from 0 to 1.

y = 2t²

Then dx is the derivative of x, times dt,

dx = 1 dt

and dy the same for y.

dy = 4t dt

Here is the path on the grid.

It goes from (0, 0) to (1, 2).

The blue arrows show F at a few points on the path.

2 Put the path in

Put x and y in t into F: P = 2t and Q = 2t².

F = ⟨2t, 2t²⟩

The line integral is P dx plus Q dy, added along the path.

∫ P dx + Q dy

So P dx + Q dy is 2t times 1, plus 2t² times 4t, all times dt.

P dx + Q dy = 2t × 1 dt + 2t² × 4t dt

Multiply out and add the like terms: 2t + 8t³.

P dx + Q dy = (2t + 8t³) dt

Careful!

Many people put t into F but forget dx and dy.

They carry the derivatives 1 and 4t.

Careful: dx and dy become dt pieces too

3 Integrate

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 1 of (2t + 8t³) dt = [t² + 2t⁴] from 0 to 1

Put in the top t = 1, then take away the bottom t = 0.

∫ from 0 to 1 of (2t + 8t³) dt = ((1)² + 2(1)⁴) − ((0)² + 2(0)⁴)

The bottom part is 0.

So it is 3.

∫ from 0 to 1 of (2t + 8t³) dt = 3

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = x² + y² over 2.

∂P/∂y = 0 = ∂Q/∂x: f = x² + y² over 2

Then the answer is f at the end minus f at the start: 3.

The same.

Right!

f(1, 2) − f(0, 0) = 3 − 0 = 3

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer