Question

Can someone help with these two questions on my Calc 3 hw? It is due tonight. Thank you!

Answer 1

Question #14:

Find the work done by the force field,
`\vec F(x,y,z)= < x-y^2, \ y-z^2, \ z-x^2 >`,

on a particle that moves along the line segment from (0,0,1) to (3,1,0).

The parametrized form of a line is given as,

`x=x_0+v_xt`

`y=y_0+v_yt`

`z=z_0+v_zt`

Where
`(x_0,y_0,z_0)` is a point the line passes through and
`\vec v= < v_x,v_y,v_z >` is the direction of the line.

`\Longrightarrow \vec v= < 3-0,1-0,0-1 > \Longrightarrow \boxed{ \vec v= < 3,1,-1 > }`

`\Longrightarrow \boxed{(x_0,y_0,z_0)=(0,0,1)}`

`z=-t+1, 0\leq t\leq 1`

This would imply that,
`dx=3dt,dy=dt,dz=-dt`

`Work, W=\int\limits^a_b {\vec F(x,y,z) \cdot} < dx,dy,dz > \,dt`

`\Longrightarrow \vec F(x,y,z)= < 3t-t^2, \ t-(-t+1)^2, \ -t+1-(3t)^2 >`

`\Longrightarrow \boxed{\vec F(x,y,z)= < 3t-t^2, \ -t^2+3t-1, \ -9t^2-t+1 > }`

`\Longrightarrow W=\int\limits^1_0 { [ < 3t-t^2, \ -t^2+3t-1, \ -9t^2-t+1 > \cdot} < 3,1,-1 > ] \,dt`

`\Longrightarrow W=\int\limits^1_0 { [( 3t-t^2)(3)+(-t^2+3t-1)(1)+(-9t^2-t+1)(-1) ] \,dt`

`\Longrightarrow W=\int\limits^1_0 { [ 9t-3t^2-t^2+3t-1+9t^2+t-1 ] \,dt`

`\Longrightarrow W=\int\limits^1_0 { [5t^2+13t-2 ] \,dt`

Using the power rule to integrate:
`(d)/(dx)[x^n]=nx^(n-1)`

`\Longrightarrow W=(5)/(3)t^3+(13)/(2)t^2-2t \left. \right|_(0)^1`

`\Longrightarrow W=[(5)/(3)(1)^3+(13)/(2)(1)^2-2(1)]-[0]`

`\Longrightarrow W=(5)/(3)+(13)/(2)-2`

`\Longrightarrow \boxed{ W=(37)/(6)} \therefore Sol.`

Question #15:

Given:

`w_m=170 \ lb`

`w_c=25 \ lb`

`w_m+w_c=195 \ lbs`

`r_s=25 \ ft`

`h_s=90 \ ft`

`1 \ rev. =2\pi \Rightarrow 3 \ rev. =\bold{6\pi}`

Find:

`W= \ ?? \ ft-lbs`

Equation:

`W=\int\ {\vec F \cdot } \, d\vec r \Longrightarrow W=\int\ {[\vec F(\vec r(t)) \cdot r'(t)}]dt`

`\vec F(x,y,z)= < 0,0,195 > \ and \ r(t)= < 25cos(t),25sin(t),(90)/(6\pi)t >`

`r'(t)= < -25sin(t),25cos(t),(15)/(\pi) >`

`\Longrightarrow W=\int\ {[\vec F(\vec r(t)) \cdot r'(t)}]dt`

`\Longrightarrow W=\int\ {[ < 0,0,195 > \cdot < -25sin(t),25cos(t),(15)/(\pi) > ]dt`

`\Longrightarrow W=\int\ {[ (195)((15)/(\pi)) ]dt`

`\Longrightarrow W=\int\ {(2925)/(\pi) dt`

Limits:
`0\leq t\leq 6\pi`

`\Longrightarrow W=\int\limits^(6\pi)_0 {(2925)/(\pi) } \, dt`

`\Longrightarrow W= {(2925)/(\pi)t \left. \right|_(0)^(6\pi)`

`\Longrightarrow W= [{(2925)/(\pi)(6\pi)]-[0]`

`\Longrightarrow \boxed{W= 17,550 \ ft-lbs} \therefore Sol.`

Let me know if these were correct! As I strive to give the most accurate answers! Thank you.