Answer:
y = (5/4)2^x
Step-by-step explanation:
The function value increases by a factor of 40/10 = 4 when x increases by 2. The function can be written as ...
y = (reference value)·(growth factor)^((x -reference)/(change in x for growth factor))
y = 10·4^((x-3)/2) . . . . . . using point (3, 10) as a reference
This can be simplified to ...
y = 10·2^(x -3) = 10/8·2^x
y = (5/4)2^x
We want to find such that
. This means
Integrating both sides of the latter equation with respect to tells us
and differentiating with respect to gives
Integrating both sides with respect to gives
Then
and differentiating both sides with respect to gives
So the scalar potential function is
By the fundamental theorem of calculus, the work done by along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it
) in part (a) is
and does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them and
) of the given path. Using the fundamental theorem makes this trivial: