Work done by F = <y, x> along y = x^2 + 1 from (0, 1) to (2, 5) Step by Step

Step by step

The question

Question: find the work done by F = ⟨y, x⟩ along y = x² + 1 from (0, 1) to (2, 5).

F = ⟨y, x⟩

along y = x² + 1 from (0, 1) to (2, 5)

1 Write the path in t

On the curve, let x be t.

Then y is t² + 1.

t goes from 0 to 2.

x = t, y = t² + 1

So x = t

x = t

and y = t² + 1, with t from 0 to 2.

y = t² + 1

Then dx is the derivative of x, times dt,

dx = 1 dt

and dy the same for y.

dy = 2t dt

Here is the path on the grid.

It goes from (0, 1) to (2, 5).

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 = t² + 1 and Q = t.

F = ⟨t² + 1, t⟩

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

∫ P dx + Q dy

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

P dx + Q dy = (t² + 1) × 1 dt + t × 2t dt

Multiply out and add the like terms: 3t² + 1.

P dx + Q dy = (3t² + 1) dt

Careful!

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

They carry the derivatives 1 and 2t.

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 2 of (3t² + 1) dt = [t³ + t] from 0 to 2

Put in the top t = 2, then take away the bottom t = 0.

∫ from 0 to 2 of (3t² + 1) dt = ((2)³ + (2)) − ((0)³ + (0))

The bottom part is 0.

So it is 10.

∫ from 0 to 2 of (3t² + 1) dt = 10

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = xy.

∂P/∂y = 1 = ∂Q/∂x: f = xy

Then the answer is f at the end minus f at the start: 10.

The same.

Right!

f(2, 5) − f(0, 1) = 10 − 0 = 10

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer