Science:Math Exam Resources/Courses/MATH101/April 2007/Question 05/Solution 1

Our first step will be to extend the cup to look like a cone and label everything as shown in the diagram.

Now, we can compute the value of y quickly via similar triangles (the big triangle and the one with side length 3),

${\frac {20+y}{6}}={\frac {y}{3}}$

and this gives y = 20cm. However, notice in the question that there are several different units everywhere and so for consistency we should use a single unit (metres). As shown, let $x_{i}^{*}$ be a sample point between 0 and 0.2 (0 to 20cm), measured from the top of the cup. Then, at this point, the radius, labeled as $r_{i}$ is

${\frac {r_{i}}{0.4-x_{i}^{*}}}={\frac {0.06}{0.4}}={\frac {3}{20}}$

and isolating gives

$r_{i}={\frac {3(0.4-x_{i}^{*})}{20}}={\frac {1.2-3x_{i}^{*}}{20}}$

Now, using the volume of a cylinder formula, we have

$V_{i}=\pi r_{i}^{2}\Delta x=\pi {\bigg (}{\frac {1.2-3x_{i}^{*}}{20}}{\bigg )}^{2}\Delta x$

Using the mass formula with $\delta$ as density, we have

$m_{i}=\delta V_{i}=1000\pi {\bigg (}{\frac {1.2-3x_{i}^{*}}{20}}{\bigg )}^{2}\Delta x$

Using the force formula, we have

$F_{i}=m_{i}g=(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{*}}{20}}{\bigg )}^{2}\Delta x$

and finally, using the work formula, we can compute that the work done at the sample point is

$W_{i}=F_{i}d=(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{*}}{20}}{\bigg )}^{2}\Delta x(0.1+x_{i}^{*})$

where we added 0.1m (10cm) to the distance d above since we are using a straw that is 10 cm above the cup. Hence, summing all these pieces gives

${\begin{aligned}\lim _{n\to \infty }\sum _{i=1}^{n}W_{i}&=\lim _{n\to \infty }\sum _{i=1}^{n}(9.8)1000\pi {\bigg (}{\frac {1.2-3x_{i}^{*}}{20}}{\bigg )}^{2}\Delta x(0.1+x_{i}^{*})\\&=\int _{0}^{20}(9.8)1000\pi {\bigg (}{\frac {1.2-3x}{20}}{\bigg )}^{2}(0.1+x)\,dx\end{aligned}}$

completing the question (we do not need to evaluate this integral).