Question

A machine is set up to cut metal strips of varying lengths and widths based on the time (t) in minutes. The change in length is given by the function I(t)=t^2-squrt(t),, and the change in width is given by w(t)=t^2-2t^1/2. Which function gives the change in area of the metal strips?

Answer 1

'a(t)=t^4-3t^(5/2)+2t'

Answer 2

If your choices are the following:
a. a(t) = t^4 + 2t
b. a(t) = t^4 + 2t + 3t^(5/2)
c. a(t) = t^4 - 3t^(5/2) + 2t`
d. a(t) = t^4 + 2t - 2t^(1/2) + sqrt(t)

The answer is letter c.a(t) = t^4 - 3t^(5/2) + 2t`
Solution:
Area= length times width
Then to get the time, a=a(t), l=l(t), and w=w(t)
So, if Area= l times width
a(t)= l(t) • w(t)
a(t)=[t^2-squrt(t)] • [t^2-2t^1/2]
a(t) = [t^2 - t^(1/2)] • [t^2 - 2t^(1/2)]
=t^4 - 3t^(5/2) + 2t