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IB MAI HL Calculus Topic 5.1 Differentiation (id: eb19323db)

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admin 发表于 2024-2-27 22:02:09 | 显示全部楼层 |阅读模式
本题目来源于试卷: IB MAI HL Calculus Topic 5.1 Differentiation,类别为 IB数学

[填空题]
The diagram below shows the cross-sectional area of a mound of beach san ml )9ajc -3ytqad(r7vd creatc7v hf0i h/,cxeszz31 ed after a high fsiez c 7vhz30 c,1/hxtide.

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

$\int_{p}^{q} \frac{x^{2}(90-x)}{1800} \mathrm{~d} x-R$

5. Determine the values of p, q and R .

R=  .




参考答案:
空格1: 30空格2: 3空格3: 2空格4: 60空格5: 3037.5空格6: 82空格7: 30空格8: 1560


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