Question

Establish a relation between the height \(h\) of the juice in the cup and radius \(r\) of the surface of the juice in the cup, if the semi-vertical angle of the cone is \(\alpha\).

Answer 1

Concept:
• In a cone, the semi-vertical angle \(\alpha\) is the angle between the axis of the cone and its generators.
• At any point in time, the volume of liquid in a cone forms a smaller similar cone.
• The ratio of the radius to the height in these similar triangles remains constant and is equal to \(\tan \alpha\).

Step 1: Identify the geometry of the cone
The cup is a right circular cone with height \(H = 15 \text{ cm}\) and base radius \(R = 5 \text{ cm}\).
Let \(\alpha\) be the semi-vertical angle of the cup.
From the right-angled triangle formed by the radius and height of the cup:
\[ \tan \alpha = \frac{\text{Radius}}{\text{Height}} = \frac{R}{H} = \frac{5}{15} = \frac{1}{3} \]

Step 2: Relate height \(h\) and radius \(r\) of the juice
At any instant, let the juice in the cup have a height \(h\) and a surface radius \(r\).
The juice forms a smaller cone inside the cup, which is similar to the cup itself.
The semi-vertical angle for the juice cone is also \(\alpha\).
From the geometry of the smaller cone:
\[ \tan \alpha = \frac{r}{h} \]

Step 3: Establish the final relation
From the results above, we can write:
\[ \frac{r}{h} = \tan \alpha \] \[ r = h \tan \alpha \] Using the numerical value \(\tan \alpha = \frac{1}{3}\):
\[ \frac{r}{h} = \frac{1}{3} \implies h = 3r \] This is the required relationship between the radius and the height of the juice.

Answer 2

Concept:
• Volume of a cone is \(V = \frac{1}{3}\pi r^2 h\).
• Rate of change of volume is given by \(\frac{dV}{dt}\).
• We use the chain rule to relate \(\frac{dV}{dt}\) and \(\frac{dh}{dt}\).
• From sub-part (i), we know \(r = \frac{h}{3}\).

Step 1: Express volume as a function of height only
Given the relation \(r = \frac{h}{3}\), substitute this into the volume formula:
\[ V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h \] \[ V = \frac{1}{3}\pi \left(\frac{h^2}{9}\right) h = \frac{\pi h^3}{27} \]

Step 2: Differentiate the volume with respect to time
Using the chain rule:
\[ \frac{dV}{dt} = \frac{d}{dh}\left(\frac{\pi h^3}{27}\right) \cdot \frac{dh}{dt} \] \[ \frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \cdot \frac{dh}{dt} \] \[ \frac{dV}{dt} = \frac{\pi h^2}{9} \frac{dh}{dt} \]

Step 3: Substitute given values to find \(\frac{dh}{dt}\)
We are given \(\frac{dV}{dt} = 0.1 \text{ cm}^3/\text{s}\) and we need to find \(\frac{dh}{dt}\) when \(h = 6 \text{ cm}\):
\[ 0.1 = \frac{\pi (6)^2}{9} \frac{dh}{dt} \] \[ 0.1 = \frac{36\pi}{9} \frac{dh}{dt} \] \[ 0.1 = 4\pi \frac{dh}{dt} \] \[ \frac{dh}{dt} = \frac{0.1}{4\pi} = \frac{1}{40\pi} \text{ cm/s} \] Thus, the juice level is rising at a rate of \(\frac{1}{40\pi} \text{ cm/s}\).

Answer 3

Concept:
• Upper surface area of juice in a conical cup is a circle: \(A = \pi r^2\).
• We know \(r = \frac{h}{3}\) from the cone's geometry.
• We use the rate \(\frac{dh}{dt} = \frac{1}{40\pi} \text{ cm/s}\) calculated in the previous step.

Step 1: Express surface area as a function of height
Substitute \(r = \frac{h}{3}\) into the area formula:
\[ A = \pi \left(\frac{h}{3}\right)^2 = \frac{\pi h^2}{9} \]

Step 2: Differentiate surface area with respect to time
Using the chain rule:
\[ \frac{dA}{dt} = \frac{d}{dh}\left(\frac{\pi h^2}{9}\right) \cdot \frac{dh}{dt} \] \[ \frac{dA}{dt} = \frac{2\pi h}{9} \cdot \frac{dh}{dt} \]

Step 3: Calculate the rate of change at \(h = 6 \text{ cm}\)
Substitute \(h = 6\) and \(\frac{dh}{dt} = \frac{1}{40\pi}\):
\[ \frac{dA}{dt} = \frac{2\pi (6)}{9} \cdot \frac{1}{40\pi} \] \[ \frac{dA}{dt} = \frac{12\pi}{9} \cdot \frac{1}{40\pi} \] \[ \frac{dA}{dt} = \frac{4}{3} \cdot \frac{1}{40} = \frac{1}{30} \text{ cm}^2/\text{s} \] Thus, the upper surface area is increasing at a rate of \(\frac{1}{30} \text{ cm}^2/\text{s}\).

Answer 4

Concept:
• Wetted surface area of a cone is its curved surface area: \(S = \pi r l\).
• Slant height \(l\) is related to radius \(r\) and height \(h\) by \(l = \sqrt{r^2 + h^2}\).
• From sub-part (i), we use \(r = \frac{h}{3}\) and from (ii), \(\frac{dh}{dt} = \frac{1}{40\pi}\).

Step 1: Express slant height \(l\) in terms of height \(h\)
Substitute \(r = \frac{h}{3}\) into the slant height formula:
\[ l = \sqrt{\left(\frac{h}{3}\right)^2 + h^2} = \sqrt{\frac{h^2}{9} + h^2} \] \[ l = \sqrt{\frac{10h^2}{9}} = \frac{\sqrt{10}}{3} h \]

Step 2: Express wetted surface area \(S\) in terms of \(h\)
Substitute \(r = \frac{h}{3}\) and \(l = \frac{\sqrt{10}}{3} h\) into \(S = \pi r l\):
\[ S = \pi \left(\frac{h}{3}\right) \left(\frac{\sqrt{10}}{3} h\right) \] \[ S = \frac{\pi \sqrt{10}}{9} h^2 \]

Step 3: Differentiate \(S\) and calculate the rate
Differentiate with respect to time using the chain rule:
\[ \frac{dS}{dt} = \frac{d}{dh}\left(\frac{\pi \sqrt{10}}{9} h^2\right) \cdot \frac{dh}{dt} = \frac{2\pi \sqrt{10} h}{9} \cdot \frac{dh}{dt} \] Substitute \(h = 6\) and \(\frac{dh}{dt} = \frac{1}{40\pi}\):
\[ \frac{dS}{dt} = \frac{2\pi \sqrt{10} (6)}{9} \cdot \frac{1}{40\pi} \] \[ \frac{dS}{dt} = \frac{12\pi \sqrt{10}}{9 \cdot 40\pi} = \frac{4 \sqrt{10}}{3 \cdot 40} = \frac{\sqrt{10}}{30} \text{ cm}^2/\text{s} \] Thus, the wetted surface area is increasing at a rate of \(\frac{\sqrt{10}}{30} \text{ cm}^2/\text{s}\).