Question

Calculate the work done by the force F along the path C.
1) F=- 9xi -5x^3y^2j +(-9z-4y^2)k ; the path is C1 U C2 U C3 where C1 is the straight line from (0,0,0) to (1,0,0), C2 is the straight line from (1,0,0) to (1,1,0), and C3 is the straight line from (1,1,0) to (1,1,1)

Answer 1

C1 can be parameterized by \(\gamma_1:[0,1]\to\mathbb{R}^3\) such that \(\mathbf{\gamma_1}(t)=\langle t,0,0\rangle\)
similarly C2 can be parameterized via \(\gamma_2(t)=\langle1,t,0\rangle\) and C3 via \(\gamma_3(t)=\langle1,1,t\rangle\)

Answer 2

the work done is computed via a line integral:
$$W=\int_{\gamma} F\ dr=\int F(\gamma(t))\cdot \gamma'(t)\ dt$$

Answer 3

for the various bits, (C1) dy & dz = 0, (C2) dx & dz = 0 & (C3) dx & dy = 0

just do the path integrals 1 by 1 and it should be pretty simple.

Answer 4

we'll take the work over each curve separately and add together:
$$W_1=\int_0^1 F(\gamma_1(t))\cdot\gamma_1'(t)\ dt$$here we have \(\gamma_1(t)=\langle t,0,0\rangle\) so \(\gamma_1'(t)=\langle1,0,0\rangle\) and we can see \(F(\gamma_1(t))=F(t,0,0)=\langle-9t,0,0\rangle\). we then see \(F(\gamma_1(t))\cdot\gamma_1'(t)=-9t\) so our integral reduces: $$W_1=\int_0^1(-9t) \ dt=\left[-\frac92t^2\right]_0^1=-\frac92$$

Answer 5

Why the answer is negative? isnt it supposed to be the work?

Answer 6

first, work is just a method of transfer of energy, like heat.

second, when a ball drops towards the earth, gravity has done some "work". but the gravitational energy of the ball is said to be zero at infinity to make the math workable. meaning that it is acquiring negative energy. it's just numbers.

third, my problem with the solution to question is that you have parameterised an integral that simply didn't need to be parameterised, and that was much simpler to start with. i will post a solution later.

Answer 7

okay

Answer 8

can you do this?

\[\int\limits_{x = 0}^{1} -9x \ dx + \int\limits_{y = 0}^{1} -5y^2 dy + \int\limits_{z = 0}^{1} -9z -4 \ dz\]

that should give you the answer you need.

\[W = \int\limits \vec F \ \bullet \vec dr = \int\limits <F_x, F_y, F_x> \ \bullet <dx,dy,dz> \\= \int\limits F_x dx + \int\limits F_y dy + \int\limits F_x dz \]

Answer 9

Yeah i can do this but i didnt know that we could do this.

Answer 10

i thought the work only has to do with the parametric equation not linear

Answer 11

can we also parametrize it and get the same answer?

Answer 12

yep, @oldrin.bataku was doing that approach.

if you want to parameterise, go for it.

i just don't see why.

@freckles

Answer 13

okay thank you

Answer 14

@pape work can be negative -- it simply means the motion is partially in the direction opposite of the force

Answer 15

for example, the work done by gravity in the *lifting* of an object is negative, which corresponds to an increase in potential energy