(a) In oder to find the range of the slope for which the line lies above in the first quadrant, you need to realize that the "worst" situation is that the line is tangent to at point , the slope corresponding to this situation is just the slope of the tangent line to graph at point , which is . Hence for any , at least part of the line will below the curve of . So the range of for which the line lies above in the first quadrant is .
(b) Given the line passes point , we can find that , so our line equation becomes . Realize that since , we know that the line must lie above the curve of . Note that the area of is just the area bounded by and from to . So the area can be expressed by the following integral:
(c) This part is similar to part (b) except we don't know the exact value of , again let's express the area bounded by and from to through the following integral:
(d) Using the expression of obtained from part (c), we can easily answer this part. In order to find the rate of change of , let's take derivative with respect to time to both sides of . We have
Here we use the given , so the area is changing at the rate of square units per second and since it is positive, we can conclude that the area is increasing.