Find the work done when the force ̰ = (−)̰+ (4^2)̰+ (2^2)̰ moves its point of application along the straight line joining (0,0,0) to (5,2,−1)
In the answers it saids "Also, (0,0,0) corresponds to = 0, whilst (5,2, −1) corresponds to = 1."
So I understand you must find the integral and such, but I don't understand how the bounds equal b=1 and a=0 from the points A and B, how did it get narrowed down to 1 and 0 from each point?
Another question I found that is similar
Determine a tangent vector and the equation of the tangent line to the space curve
r(t) = (t)i+ (t^2)j+ (t^3)k
at the point (2, 4, 8)
Differentiating once with respect to t gives r˙(t) = i+(2t)j+(3t^2)k
so that a tangent vector at (2, 4, 8), where t = 2, is i+4j+12k.
How was is it concluded that t=2?
Given point $A$ and $B$, a common way of parametrization of the straight line connecting $A$ and $B$ is $r(t)=(1-t)A+tB$.
For this particular choice of parametrization, we can easily see that $r(0)=A$ adn $r(1)=B$.
For your second question, the parametrization has been chosen to be $(t, t^2, t^3)$.
To find the corresponding $t$ that gives point $(2,4,8)$. We solve for
$$(t,t^2, t^3)=(2,4,8)$$
and we can see from the first component that $t=2$. The other two components hold as well.