I need help on calc problem?
An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 39 m per year at a time when the region is 480 m wide. How fast is the area changing at that point in time?
- stanschimLv 73 weeks ago
A = lw = (2w)(w) = 2w^2
Differentiating with respect to t gives:
A'(t) = 4ww'
A'(t) = 4(480)(39) = 74880 square meters per year
- 3 weeks ago
Assume the growth is constrained so that the length is always twice the width: l = 2w
The zone is rectangular, so A = lw = (2w)w = 2w^2
We want to know the rate of change of the area over time, so take derivatives with respect to time on both sides: A' = 4w * w'. The w' on the right comes from chain rule.
We're given a width and a fixed rate of change of the width in the problem, so we can plug those into this equation to get the rate of change of the area:
A' = 4(480 m)(39 m/year) = 74880 m^2/year
Hope this helps!