Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t))i + (1 − cos(t))j, 0 ≤ t ≤ 2π.

By direct calculation we will find that the work done is equal to 2π²

The formula to compute the work done is given by:

Here we have:

This means that:

And we know that 0 ≤ t ≤ 2π, so b = 0 and a = 2π

Replacing that in the work integral we get:

the sin(t) integral can be removed because it is equal to zero, so we get:

If you want to learn more, you can read:

link

Integrate the force field along the given path (call it C):

neither i nor ii 56 / 4.9 = 11.91489362 and 82 / (11+5) =5.125 so i is wrong

ii. (24/8+2) times 3 = 15 and (42+9)/2 =25.5

step-by-step explanation:

56

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb