# Thread: Applications of differentation #3

1. ## Applications of differentation #3

1) Show $y =\frac{-x^2}{2700}(x-90)$
2) North South distance d-metre between the south semi circle boundary and the cubic model of the wall can be found using:
$d = 19.5-\sqrt{6.25-(x-25)^2} + \frac{x^2}{2700}(x-900)$
a) Domain of $d$ $x$ $\varepsilon [x_{1}, x_{2}]$. Find $x_{1}$ and $x_{2}$

I have attempted these and I am stuck. Any help/hints will be appreciated.

2. Originally Posted by Shiraz
1) Show $y =\frac{-x^2}{2700}(x-90)$
This makes no sense at all. I can assume that the graph of y= f(x) is the curve that, in your picture, looks like two straight lines but you haven't specified any coordinate system. Where are x and y measured from? If we assume that (0, 0) is at the point on the left where the curves seems to end and that the x and y components are the distances in meters we still have only that y(0)= 0 and that y(60)= 40. What reason do we have to think that the curved wall comes back to the base so that y(90)= 0?

2) North South distance d-metre between the south semi circle boundary and the cubic model of the wall can be found using:
$d = 19.5-\sqrt{6.25-(x-25)^2} + \frac{x^2}{2700}(x-900)$
a) Domain of $d$ $x$ $\varepsilon [x_{1}, x_{2}]$. Find $x_{1}$ and $x_{2}$

I have attempted these and I am stuck. Any help/hints will be appreciated.

3. Originally Posted by Shiraz
1) Show $y =\frac{-x^2}{2700}(x-90)$
2) North South distance d-metre between the south semi circle boundary and the cubic model of the wall can be found using:
$d = 19.5-\sqrt{6.25-(x-25)^2} + \frac{x^2}{2700}(x-900)$
a) Domain of $d$ $x$ $\varepsilon [x_{1}, x_{2}]$. Find $x_{1}$ and $x_{2}$

I have attempted these and I am stuck. Any help/hints will be appreciated.
If I were you I would repost this question with a much more thorough explaination of what you mean, its very vague and just downright confusing. If you post back this question with better explainations and all the information I'm sure you'll get some help with them.