7499
/www/nrich/html/content/id/7499/
1
2011-03-21T17:56:35
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<ul id="stemLinks">
<li><a href="http://nrich.maths.org/7419">Warm-up</a></li>
<li><a href="http://nrich.maths.org/6439">Try this next</a></li>
<li><a href="http://nrich.maths.org/6426">Think higher</a></li>
<li><a href="http://en.wikipedia.org/wiki/Solid_of_revolution">Read: mathematics</a></li>
<li><a href="http://en.wikipedia.org/wiki/Dam">Read: science</a></li>
<li><a href="http://plus.maths.org/content/puzzle-page-24">Explore further</a></li>
</ul>
<div> </div>
<br></br>
<p>In the problem <a href="http://nrich.maths.org/7419&part=">Fill Me Up</a>, we invited you to sketch graphs showing the height of water against the volume for six containers, as they filled up.<br></br>
<br></br>
Instead of sketching graphs or producing experimental plots, we could produce graphs by analysing the shape of each container and working out the function linking height and volume. Some functions are easier to work out analytically than others!<br></br>
</p>
<table style="" border="1">
<tbody>
<tr>
<td> <mdo:image alt="" height="300" src="pint%20glass.jpg" width="200"></mdo:image></td>
<td> <mdo:image alt="" height="300" src="conical%20flask.jpg" width="200"></mdo:image></td>
</tr>
</tbody>
</table>
<br></br>
<p>The Pint Glass and the lower portion of the Conical Flask are both <a href="http://en.wikipedia.org/wiki/Frustum">frustums</a> of cones.<br></br>
We can use what we know about the volume properties of a cone to help us to analyse what happens when we fill the pint glass or conical flask with water.<br></br>
<br></br>
Imagine a cone sitting on its point being filled with water:<br></br>
<mdo:image alt="" height="250" src="cone.jpg" width="500"></mdo:image><br></br>
In the second picture, the height of the water level has doubled. How has the volume of water changed?<br></br>
What if the height had trebled?<br></br>
What if the height had increased by a factor of $n$?<br></br>
<br></br>
How would I need to increase the height in order to double the volume?<br></br>
How would I need to increase the height in order to treble the volume?<br></br>
How would I need to increase the height in order to increase the volume by a factor of $n$?<br></br>
<br></br>
What would a graph of volume (y) against height (x) look like?<br></br>
What would a graph of height (y) against volume (x) look like?<br></br>
<br></br>
<span style="font-weight: bold;">The Pint Glass is not a whole cone, it is a frustum. How could you use the graph for a cone to work out what the graph for the Pint Glass would look like?</span><br></br>
<br></br>
<span style="font-weight: bold;">Extension challenge</span><br></br>
Using a similar analysis, can you work out the shape of the graph for height against volume for a cone sitting on its base rather than its point?<br></br>
Can you use your graph to work out what the graph for the Conical Flask would look like?<br></br>
<br></br>
Very challenging extension:<br></br>
Can you work out an analytical form for the function linking volume and height for a spherical flask?<br></br>
</p>
<div class="framework"><br></br>
<span style="font-weight: bold;">Pictures</span><br></br>
<br></br>
http://commons.wikimedia.org/wiki/File:Pyrex_Conical_Flask.jpg<br></br>
http://commons.wikimedia.org/wiki/File:Pint_glass_300x509.jpg</div>
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<mdoxml version="1.0"><p><span class="editorial">Well done to those of you who submitted solutions - this question is harder than it looks!</span></p>
<p><span class="editorial">Tom from the Norwich School, gave some good explanations of the first section:</span></p>
<p>We know that the volume of a cone is given by $V_1=\frac {\pi}{3} r^2 h$.</p>
<p>1) We need to calculate the relationship between the height of the cone and the radius of its base.</p>
<p>Suppose we increase the height of our cone, and keep the proportions, i.e. the ratio of height to the radius, the same. Let $y$ be the height of our new cone, and $x$ be the radius. The gradient, $\frac{h}{r}$, is therefore constant, so the relationship between $y$ and $x$ is given by:<br></br>
$$ \begin{align*} y &= \frac{h}{r}x \\ \text{If we set } y=2h &\Rightarrow 2h=\frac{h}{r}x \\\Rightarrow x& =2r \end{align*} $$</p>
<p style="text-align: center;"><mdo:image alt="" src="7499%20-%20picture%201.jpg" style="width: 430px; height: 182px;"></mdo:image></p>
<p>From this, we are able to find the new volume:$$V_2 = 2^3 \bigg(\frac {\pi}{3} r^2 h \bigg)=2^3 V_1 $$</p>
<p>2) If the height had trebled, then we use the same method:</p>
<p>$$ \begin{align*} y &= \frac{h}{r}x \\ \text{If we set } y=3h &\Rightarrow 3h=\frac{h}{r}x \\\Rightarrow x& =3r \end{align*} $$ and so $$V_2 = 3^3V_1$$ </p>
<p>3) Noticing a pattern, we see that if the height had increased by a factor of <em>n</em>, then $$V_n = n^3 \bigg(\frac {\pi}{3} r^2 h \bigg)=n^3 V_1 $$</p>
<p>4) Suppose we want to double the volume by changing the height (not forgetting the radius changes too!)i.e. $$V_2 = \frac {\pi}{3} r_2^2 h_2 = 2V_1 = \frac {\pi}{3} r_1^2 h_1$$</p>
<p>If $h_2 = kh_1$, then we know the radius also increases like $r_2 = kr_1$. Therefore $$\begin{align*}V_2 = k^3V_1 \Rightarrow k &= \sqrt[3]{2} \\\text{ i.e. } h_2 &= \sqrt[3]{2}h_1 \end{align*}$$ Similarly, if we want to treble the volume, we should increase the height by a factor of $\sqrt[3]{3}$. In general, if we want to increase the volume by a factor of n, we should increase the
height by a factor of $\sqrt[3]{n}$. </p>
<p><br></br>
<span class="editorial">Cecil sent us graphs that demonstrated the cubic relationship:</span><br></br>
5) The blue line is a graph of volume against height, and the red line is a graph of height against volume. We can see they're reflections of each other in the line $v=h$. </p>
<p style="text-align: center;"><mdo:image alt="" src="fill%20me%20up.jpg" style="width: 282px; height: 240px;"></mdo:image></p>
<p> </p>
<p><span class="editorial">Extension Challenge: Martin from All Saints School wrote to tell us this explanation about the graph for a pint glass.</span><br></br>
<br></br>
The graph of a pint glass will have a steeper gradient at the start than that of the cone as the volume will rise more quickly and the height more slowly because there is a larger surface area at the bottom. However, the total volume and total height will be smaller than the cone.<br></br>
</p>
<p><span class="editorial">Robert from Bishop Tonnos High School, Canada calculated the volume of a conical flask.</span></p>
<p>If you remove a cone of height h from the top of a larger cone (with the same proportions) of height H, the volume of the remaining object, a frustum, is $V = \frac{1}{3}\pi\tan^2{\theta}(H^3 - h^3)$ where $\theta$ is the angle between the vertical and the side. (Trigonometry implies $\tan(\theta) = \frac{H}{R}$, where $R$ is the radius of the large cone.) This is tricky to calculate, as we
can't measure $H$ or $h$ easily. Robert noticed that $$\tan(\theta) = \frac{R-r}{H-h}$$ where r in the radius of the small cone. </p>
<p style="text-align: center;"><mdo:image alt="" src="frustram2.png" style="width: 235px; height: 118px;"></mdo:image></p>
<p style="text-align: left;">Denote the height of the frustam $H_f$ and the height of the cylindrical section $H_c$. As we're now considering part of a cone standing on its base, let $h$ be the level of the water. (NB $h$ is equivalent to $H-h$ in the previous formulae as it's upside down!) the The formula for the amount of water is the vessel is:</p>
<p style="text-align: left;">$$ V = \left\{\begin{array}{l l}\frac{1}{3} \pi\frac{(R-r)^2}{H_f^2}(H_f^3 - (H_f-h)^3) & \quad h\leq H_f\\\frac{1}{3} \pi\frac{(R-r)^2}{H_f} + \pi r^2(h-H_f) & \quad h > H_f \\ \end{array} \right.$$</p>
<p style="text-align: left;">where the first term in the second equation is the frustum term and the second is the cylinder term. </p>
<p style="text-align: left;"><strong>Spherical vessel:</strong></p>
<p style="text-align: left;"><span class="editorial">To derive the volume of a spherical vessel requires integration, which you'll learn at A-level, so don't worry if you don't understand this, it's getting really complicated!</span></p>
<p style="text-align: left;">The volume of a <a href="http://mathworld.wolfram.com/SphericalCap.html">spherical cap</a> cut from a sphere of radius $a$ is given by:</p>
<p style="text-align: left;">$$V= \frac{1}{6}\pi h(3b^2+h^2)$$ where $h$ is its height and $b^2 = 2ah - h^2$. We can use this result to find the volume of a spherical vessel. Suppose our vessel is a sphere with a spherical cap of height h removed, and a cylinder of radius c attached.</p>
<p style="text-align: center;"><mdo:image alt="" src="spherical.png" style="width: 235px; height: 118px;"></mdo:image></p>
<p style="text-align: left;">We can therefore write an equation for the volume of the vessel. It will again be different depending on whether the height of the water, $w$ is greater or less than $2a - h$ (the point where the water will start filling the cylinder instead of the sphere). For $w < 2a-h$, the volume is itself a spherical cap of height w. </p>
<p style="text-align: center;">$$V_{vessel} =\left\{\begin{array}{l l} \frac{1}{6}\pi h(3b^2+w^2) & \quad w\leq 2a - h \\ S + \pi c^2(w-(2a-h) & \quad w > 2a-h \\ \end{array} \right.$$ </p>
<p style="text-align: center;">where we've written the volume of the truncated sphere as $S$, where $$S = \frac{4\pi}{3}a^3 - \frac{1}{6}\pi h(3b^2+h^2)$$</p>
<p style="text-align: center;"> </p>
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<h3>Why do this problem?</h3>
<a href="http://nrich.maths.org/7499">This problem</a> follows on from <a href="http://nrich.maths.org/7419&part=">Fill Me Up</a>, and gives students the opportunity to use volume scale factors of enlargement to work out the relationship between the volume and the height of a cone.<br></br>
<h3>Possible approach</h3>
<div>Perhaps start by asking students to sketch the graphs from the problem <a href="http://nrich.maths.org/7419&part=">Fill Me Up</a>. <a href="/content/id/7499/Fill%20Me%20Up.pdf">Here</a> is a worksheet showing the containers.</div>
<div> </div>
<div>"Imagine we wanted to plot the graphs accurately by working out the equations linking height to volume. Some parts of the containers will be easier to work out than others - which will be easiest? Which will be hardest?"</div>
<div> </div>
<div>Take time to discuss students' ideas, relating it back to the graphs sketched in the first problem.</div>
<div> </div>
<div>"Let's try to analyse how the height changes as the Pint Glass is filled." </div>
<div>"The Pint Glass can be thought of as part of a cone (a frustum), so I'd like you to consider a cone filling with water first."</div>
<div>Give students <a href="/content/id/7499/Fill%20Me%20Up%20Too.pdf">this worksheet</a> to work on in groups of 3 or 4. These <a href="http://nrich.maths.org/content/id/2290/Roles.pdf">roles</a> may be useful for students who are not used to working collaboratively on a problem. Make it clear that your expectation is for all students in the group to be able to explain their thinking clearly and
that anyone might be chosen to present the group's conclusions at the end of the lesson.</div>
<div> </div>
<div>Finally, allow time at the end of the lesson (or two lessons) for groups to present their thinking to the rest of the class. </div>
<br></br>
<h3>Key questions</h3>
<div>What happens to the volume of a cone when I enlarge it by a scale factor of 2, 3, 4, 5... k?</div>
<div>If the volume of water is $10$cm$^3$ when the height of the water is $1$cm, what will the volume be when the height is $2, 3, 4...x$cm?</div>
<div>How could this be represented graphically?</div>
<h3>Possible extension</h3>
There are two extension tasks suggested in the problem: analysing the inverted cone is a reasonably straightforward extension, but analysing the spherical flask is much much more challenging.<br></br>
<br></br>
<a href="http://nrich.maths.org/6439&part=">Immersion</a> and <a href="http://nrich.maths.org/6426&part=">Brimful</a> both offer extension possibilities for considering functional relationships relating to volume.<br></br>
<h3>Possible support</h3>
<a href="http://nrich.maths.org/6923&part=">Growing Rectangles</a> offers a good introduction to proportional relationships between length, area and volume.<br></br>
<br></br>
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Imagine a conical container that contains a volume of 10cm$^3$ when
filled to a height of 1cm:<br></br>
<br></br>
<table border="1">
<tbody>
<tr>
<td style="text-align: center;">Height</td>
<td style="text-align: center;">Volume</td>
</tr>
<tr>
<td style="text-align: center;">1</td>
<td style="text-align: center;">10</td>
</tr>
<tr>
<td style="text-align: center;">2</td>
<td style="text-align: center;">80</td>
</tr>
<tr>
<td style="text-align: center;">3</td>
<td style="text-align: center;">270</td>
</tr>
</tbody>
</table>
<br></br>
What is the relationship between the height and volume as this cone
is filled? <br></br>
How could you draw a graph to show this?<br></br>
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Fill Me Up Too
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
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