[填空题]
The diagram below shows the cross-sectional a9:q,yls;tl5h q- tuy( d8m8dtjqfhc+rea ofu9u tai6hlw 6sf2cu6i46d d 1p a mound of beach sand created after a hil uitdu4cs au1i dw9666 hp2f6gh tide.
The curve of the cross section can be modelled by the following equation
$y=\frac{x^{2}(90-x)}{1800}$
where y represents the vertical height of the mound in \mathrm{cm} and x denotes the horizontal width in \mathrm{cm} , from the start of the mound.
1. At a horizontal width of x=30 , determine
1. The vertical height of the mound at this point;y= cm.
2. The gradient of the mound curve at this point.y'=$\frac{a}{b}$;a= ,b= .
2. 1. Find the value of x which corresponds to the maximum the vertical height of the mound.
x= .
2. Hence, find the maximum vertical height of the mound.A= cm$^2$.
3. Calculate the cross-sectional area of the mound, rounding your answer to one decimal place.
A child uses a toy shovel to remove the top of sand mound, as illustrated by the line segment MN below. Point M has coordinates at (30,30) .
4. Determine the coordinates of point N .
N( , )
The cross-sectional area removed by the child can be expressed by the following integral
