Science:Math Exam Resources/Courses/MATH101/April 2009/Question 07 (b)/Solution 1

A picture is included at the bottom of the page. Let $x_{i}^{*}$ be a sample point.

First, we use similar triangles as in the diagram to see that

$\displaystyle {\frac {1}{2}}={\frac {y}{2-x_{i}^{*}}}$

Isolating yields that

$\displaystyle y=1-{\frac {x_{i}^{*}}{2}}$

Now we use hint 3 and write down the necessary formulas. For volume of the ith piece, we have

$V_{i}=y\Delta x\cdot 10=10(1-{\frac {x_{i}^{*}}{2}})\Delta x=(10-5x_{i}^{*})\Delta x$

For the mass of the ith piece, where $density=\rho =1000$, we have

$m_{i}=V\cdot \rho =(10-5x_{i}^{*})\Delta x(1000)=1000(10-5x_{i}^{*})\Delta x$

For the force, where $g=9.8$, we have

$F_{i}=m_{i}a=9.8(1000(10-5x_{i}^{*})\Delta x)=9800(10-5x_{i}^{*})\Delta x$

For the work on the ith piece, we have

$W_{i}=F_{i}\cdot d=9800(10-5x_{i}^{*})\Delta x(x_{i}^{*})=9800x_{i}^{*}(10-5x_{i}^{*})\Delta x$

and so the total work done is

$W=\lim _{n\to \infty }\sum _{i=1}^{n}W_{i}=\int _{0}^{2}9800x(10-5x)\,dx$

completing the question.