Work done by F = <x, y^2> along y = x^2 from (0, 0) to (2, 4) Step by Step

Step by step

The question

Question: find the work done by F = ⟨x, y²⟩ along y = x² from (0, 0) to (2, 4).

F = ⟨x, y²⟩

along y = x² from (0, 0) to (2, 4)

1 Write the path in t

On the curve, let x be t.

Then y is t².

t goes from 0 to 2.

x = t, y = t²

So x = t

x = t

and y = t², with t from 0 to 2.

y = t²

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, 0) to (2, 4).

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 = t and Q = t⁴.

F = ⟨t, t⁴⟩

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

∫ P dx + Q dy

So P dx + Q dy is t times 1, plus t⁴ times 2t, all times dt.

P dx + Q dy = t × 1 dt + t⁴ × 2t dt

Multiply out and add the like terms: t + 2t⁵.

P dx + Q dy = (t + 2t⁵) 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 (t + 2t⁵) dt = [t²2 + t⁶3] from 0 to 2

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

∫ from 0 to 2 of (t + 2t⁵) dt = ((2)²2 + (2)⁶3) − ((0)²2 + (0)⁶3)

The bottom part is 0.

So it is 70/3.

∫ from 0 to 2 of (t + 2t⁵) dt = 703

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = x² over 2 + y³ over 3.

∂P/∂y = 0 = ∂Q/∂x: f = x² over 2 + y³ over 3

Then the answer is f at the end minus f at the start: 70/3.

The same.

Right!

f(2, 4) − f(0, 0) = 70/3 − 0 = 70/3

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer