Work done by a radial force field...?

Question

Let F be the radial force field F = xi + yj. Find the work done by this force along the following two curves, both which go from (0, 0) to (10, 100). 

A. If C1 is the parable: x = t, y = t^2, 0 ≤ t ≤ 10, then ∫(of C1) F dr = 

B. If C2 is the straight line segment: x = 10t^2, y = 100t^2, 0 ≤ t ≤ 1, then ∫(of C2) F dr =

Answer 1

(A)

Radial force field F = xi + yj

Parabola C1  x,y coordinates are x = t , y = t²

Parabola r(t) = t i + t² j , 0 ≤ t ≤ 10

Apply derivative with respect to t on both sides.

dr(t) = (i + 2tj)dt

work done by the radial force W = ?

The differential work is dot product of Force and path length

dW = F•dr(t)

dW = (xi + yj)•(i + 2tj)dt

dW = (xi•i + 2tyj•j)dt

Substitute i•i  = j•j = 1

dW = (x + 2ty)dt

Substitute x = t , y = t²

dW = (t + 2t*t²)dt

dW = (t + 2t³)dt

Now integrate with respect to t in the interval 0 to 10

∫dW = ∫(t + 2t³)dt

W = (t²/2 + (2/4)t⁴)

W = (t² +t⁴)/2

Substitute intervals 0 to 10

W = (10² + 10⁴)/2

W = 5050 J

The work done is 5050 Joules.

Answer 2

(B)

Radial force field F = xi + yj

Parabola C1  x,y coordinates are x = 10t² , y = 100t⁴

Parabola r(t) = 10t² i + 100t⁴ j , 0 ≤ t ≤ 1

Apply derivative with respect to t on both sides.

dr(t) = 10*2t i dt + 100*4t³ j dt

dr(t) = (20t i + 400t³ j)dt

work done by the radial force W = ?

The differential work is dot product of Force and path length.

dW = F•dr(t)

dW = (xi + yj)•(20t i + 400t³ j)dt

dW = ( 20tx i•i + 400t³y j•j )dt

Substitute i•i  = j•j = 1

dW = ( 20tx + 400t³y ) dt

Substitute x = 10t² , y = 100t⁴

dW = ( 20t*10t² + 400t³*100t⁴ ) dt

dW = ( 200t³ + 40000t⁷ ) dt

Now integrate with respect to t in the interval 0 to 1

∫dW = ∫( 200t⁴/4+ 40000t⁸ /8) dt

W = ( 50t³ + 5000t⁷ )

Substitute intervals 0 to 1

W = ( 50*1³ + 5000*1⁷ )

W = ( 50+ 5000 )

W = 5050 J

The work done is 5050 Joules.