Example Question - water level
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Determining the Height of Water in an Inverted Cone
<p>Let the height of the cone be \( h \) cm, and the height of the water level when the cone is upright be \( \frac{h}{2} \) cm.</p> <p>We can use the formula for the volume of a cone, \( V = \frac{1}{3}\pi r^2 h \), where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height.</p> <p>When the cone is full, the volume is \( V_{\text{full}} = \frac{1}{3}\pi r^2 h \). When it is filled to half its height, the volume is \( V_{\text{half}} = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 \left(\frac{h}{2}\right) = \frac{1}{24}\pi r^2 h \), because the radius is also halved at half the height (due to similar triangles).</p> <p>When the cone is inverted, the same volume of water \( V_{\text{half}} \) will now fill the cone to a new height \( x \) with an unknown radius \( r' \).</p> <p>The relationship between the radii and heights due to similar triangles in the inverted cone is \( \frac{r}{h} = \frac{r'}{x} \) or \( r' = \frac{rx}{h} \).</p> <p>Equating the volume of water when the cone is upright and inverted gives us: \( \frac{1}{24}\pi r^2 h = \frac{1}{3}\pi r'^2 x = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).</p> <p>Simplifying this equation gives: \( \frac{1}{24} h = \frac{1}{3}\left(\frac{r^2 x^3}{h^2}\right) \).</p> <p>Multiplying both sides by \( h^2 \) and dividing by \( r^2 \) gives: \( \frac{h^3}{24r^2} = \frac{1}{3} x^3 \).</p> <p>Multiplying both sides by 3 and taking the cube root gives: \( x = \sqrt[3]{\frac{h^3}{8r^2}} \).</p> <p>But since \( r \) is the radius of the cone when full, and we need to express \( x \) in terms of \( h \) only, we recognize that the radius \( r \) will be eliminated in the relation because it is a proportionality constant given the shape of the cone is unchanged.</p> <p>Therefore, we simplify to: \( x = \frac{h}{2} \) (since the volume is one-eighth of the full volume, and by the cube root, the height is half).</p>
Determining the Height of the Water Level in an Inverted Cone
<p>Let the original height of the cone be \( h \) and the radius be \( r \).</p> <p>When the cone is half-filled with water, the volume \( V \) of the water is \( \frac{1}{3}\pi\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) \) because the height of the water is \( \frac{h}{2} \).</p> <p>When the cone is inverted, let \( x \) be the height of the water level in terms of \( h \), and \( r' \) be the new radius of the water level.</p> <p>By similar triangles, \( \frac{r'}{x} = \frac{r}{h} \rightarrow r' = \frac{rx}{h} \).</p> <p>The volume of the water remains constant, so \( V = \frac{1}{3}\pi r'^2 x = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).</p> <p>Equating the two expressions for \( V \), we get \( \frac{1}{3}\pi\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).</p> <p>Simplify to find \( x \):</p> <p>\( \left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right) = \left(\frac{rx}{h}\right)^2 x \).</p> <p>\( \frac{r^2 h}{4} = \frac{r^2 x^3}{h^2} \).</p> <p>Multiply both sides by \( h^2 \):</p> <p>\( h^3 = 4x^3 \).</p> <p>Take the cube root of both sides to solve for \( x \):</p> <p>\( x = \frac{h}{\sqrt[3]{4}} \).</p> <p>Therefore, the height of the water level \( x \) in terms of the original height \( h \) when the cone is inverted is \( x = \frac{h}{\sqrt[3]{4}} \).</p>
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