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: A conical cup forms a right circular cone geometric profile. When juice fills the cup up to a certain height \(h\), the liquid mass itself forms a smaller right circular cone inside the cup container. The semi-vertical angle \(\alpha\) remains entirely constant throughout the filling process because it is dictated solely by the fixed physical dimensions of the outer container structure. By looking at a vertical cross-section of the cone, we can use simple right-angled trigonometry to establish a definitive relation between the instantaneous fluid radius \(r\) and fluid height \(h\).

Step 1: Visualizing and drawing the right triangle from the cross-section.
Consider the cross-section of the cone containing the central vertical axis. Let the vertex of the cone at the bottom base point be denoted as \(O\). The vertical height axis rises straight up through the center of the cone to the upper liquid surface. Let the height of the fluid at any given point in time be represented by the segment length \(h\). The radius of the upper circular disk surface of the juice forms a horizontal straight line perpendicular to the vertical axis line; let this segment length be represented by \(r\). This geometric layout forms a clear right-angled triangle where:
• The side opposite to the semi-vertical angle \(\alpha\) is the instantaneous radius of the liquid surface, \(r\).
• The side adjacent to the semi-vertical angle \(\alpha\) along the central vertical axis line is the instantaneous height of the juice, \(h\).

Step 2: Applying the basic trigonometric tangent ratio.
By definition, the tangent function of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the adjacent side: \[ \tan \alpha = \frac{\text{Opposite Side}}{\text{Adjacent Side}} \] Substituting our precise geometric variables into this trigonometric definition gives: \[ \tan \alpha = \frac{r}{h} \]

Step 3: Rearranging the equation to isolate the radius \(r\).
To find the explicit structural relationship expressing \(r\) as a direct function of \(h\), we cross-multiply both sides of the equation by \(h\): \[ r = h \cdot \tan \alpha \] This gives us our definitive geometric constraint equation. Additionally, let us calculate the numeric value of \(\tan \alpha\) using the full physical parameters of the cup container given in the description (radius \(R = 5\text{ cm}\) and depth \(H = 15\text{ cm}\)): \[ \tan \alpha = \frac{R}{H} = \frac{5}{15} = \frac{1}{3} \] Substituting this constant value back into our derived formula gives: \[ r = \frac{h}{3} \] Thus, both \(r = h \tan \alpha\) and \(r = \frac{h}{3}\) are perfectly accurate, and \(r = h \tan \alpha\) perfectly satisfies Option (A).

Answer 2

Concept: This question deals with related rates of change under differential calculus. We are given the volumetric flow rate at which juice enters the container, which is mathematically represented as \(\frac{dV}{dt} = 0.1\text{ cm}^3\text{/s}\). We need to determine the instantaneous rate of change of the vertical fluid level height, denoted as \(\frac{dh}{dt}\), at the specific instant when \(h = 6\text{ cm}\). To do this, we write the volume formula of a cone, use our geometric relationship to express it in terms of a single variable \(h\), and differentiate it implicitly with respect to time \(t\).

Step 1: Stating the volume formula and substituting the single variable constraint.
The standard equation for the total volume \(V\) of a circular cone is given by: \[ V = \frac{1}{3}\pi r^2 h \] From our analysis in part (i), we established that the radius at any instant is directly proportional to height by the fraction: \[ r = \frac{h}{3} \] Let us substitute this expression for \(r\) into our volume equation so that the volume is written purely as a function of the single operational variable \(h\): \[ V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h \] Expand the squared term carefully: \[ V = \frac{1}{3}\pi \left(\frac{h^2}{9}\right) h \] Combine the constants in the denominator and the powers of \(h\): \[ V = \frac{\pi}{27} h^3 \]

Step 2: Differentiating the volume equation with respect to time \(t\).
Using the chain rule of differential calculus, we differentiate both sides of the structural volume equation implicitly with respect to time \(t\): \[ \frac{dV}{dt} = \frac{d}{dt}\left( \frac{\pi}{27} h^3 \right) \] \[ \frac{dV}{dt} = \frac{\pi}{27} \cdot \left( 3h^2 \cdot \frac{dh}{dt} \right) \] Simplifying the constant fractional coefficients (\(\frac{3}{27} = \frac{1}{9}\)) yields: \[ \frac{dV}{dt} = \frac{\pi h^2}{9} \cdot \frac{dh}{dt} \]

Step 3: Substituting the known instantaneous values to isolate \(\frac{dh}{dt}\).
We are given the following explicit values from the problem statement:
• \(\frac{dV}{dt} = 0.1\text{ cm}^3\text{/s} = \frac{1}{10}\text{ cm}^3\text{/s}\)
• \(h = 6\text{ cm}\) Substitute these numerical values directly into our differentiated related rates equation: \[ \frac{1}{10} = \frac{\pi \cdot (6)^2}{9} \cdot \frac{dh}{dt} \] Evaluate the square of 6: \[ 6^2 = 36 \] Substitute and simplify the fraction: \[ \frac{1}{10} = \frac{36\pi}{9} \cdot \frac{dh}{dt} \] Since \(36 \div 9 = 4\), the expression becomes: \[ \frac{1}{10} = 4\pi \cdot \frac{dh}{dt} \] Isolate the rate of height increase \(\frac{dh}{dt}\) by dividing both sides by \(4\pi\): \[ \frac{dh}{dt} = \frac{1}{10 \cdot 4\pi} = \frac{1}{40\pi}\text{ cm/s} \] Wait, let's re-verify the substitution. If \(\frac{dh}{dt} = \frac{1}{40\pi}\), let's check Option (A). Option (A) is \(\frac{1}{40\pi}\). Let's fix the correct choice tag to match Option (A).

Answer 3

Concept: The upper surface of the juice inside the cone forms a flat circular disk shape. The area \(A_s\) of this upper boundary circular surface is given by the standard geometric formula \(A_s = \pi r^2\). We need to determine the rate of change of this surface area with respect to time, written as \(\frac{dA_s}{dt}\), at the precise moment when \(h = 6\text{ cm}\). We can achieve this by expressing \(A_s\) completely in terms of \(h\) and applying differentiation principles.

Step 1: Formulating the surface area equation in terms of height \(h\).
The upper surface area is defined as: \[ A_s = \pi r^2 \] Using our known linear constraint relationship \(r = \frac{h}{3}\), we substitute for \(r\): \[ A_s = \pi \left(\frac{h}{3}\right)^2 = \frac{\pi h^2}{9} \]

Step 2: Differentiating with respect to time \(t\).
Now, we differentiate the area formula with respect to time \(t\) using implicit differentiation: \[ \frac{dA_s}{dt} = \frac{d}{dt}\left( \frac{\pi h^2}{9} \right) = \frac{\pi}{9} \cdot \left( 2h \cdot \frac{dh}{dt} \right) \] \[ \frac{dA_s}{dt} = \frac{2\pi h}{9} \cdot \frac{dh}{dt} \]

Step 3: Plugging in known instantaneous values.
From our previous calculations, we know the exact values at this instant are:
• \(h = 6\text{ cm}\)
• \(\frac{dh}{dt} = \frac{1}{40\pi}\text{ cm/s}\) Substitute these numerical measurements into our surface area rate derivative expression: \[ \frac{dA_s}{dt} = \frac{2\pi \cdot 6}{9} \cdot \left(\frac{1}{40\pi}\right) \] Let us simplify this expression step-by-step: \[ \frac{2\pi \cdot 6}{9} = \frac{12\pi}{9} = \frac{4\pi}{3} \] Now multiply this simplified fraction by the height rate of change component: \[ \frac{dA_s}{dt} = \frac{4\pi}{3} \cdot \frac{1}{40\pi} \] Notice that the \(\pi\) terms in the numerator and denominator cancel out completely: \[ \frac{dA_s}{dt} = \frac{4}{3 \cdot 40} = \frac{4}{120} = \frac{1}{30}\text{ cm}^2\text{/s} \] Wait! Let's re-verify the values. Let's look at another direct way: \[ A_s = \pi r^2 \implies \frac{dA_s}{dt} = 2\pi r \frac{dr}{dt} \] Since \(r = \frac{h}{3} = \frac{6}{3} = 2\text{ cm}\). And \(r = \frac{h}{3} \implies \frac{dr}{dt} = \frac{1}{3}\frac{dh}{dt} = \frac{1}{3}\cdot\frac{1}{40\pi} = \frac{1}{120\pi}\). Then: \[ \frac{dA_s}{dt} = 2\pi (2) \left(\frac{1}{120\pi}\right) = \frac{4\pi}{120\pi} = \frac{1}{30}\text{ cm}^2\text{/s} \] Let's make sure the options are written cleanly. Let's write the response value as \(\frac{1}{30}\text{ cm}^2\text{/s}\).

Answer 4

Concept: The wetted surface area of the cup corresponds directly to the curved surface area (\(A_c\)) of the cone formed by the liquid volume. The standard mathematical expression for the lateral curved surface area of a circular cone is given by: \[ A_c = \pi r l \] Where \(l\) represents the slant height of the cone of liquid. By the Pythagorean theorem, the slant height is related to the radius and height by the expression \(l = \sqrt{r^2 + h^2}\). We need to determine the rate of change \(\frac{dA_c}{dt}\) when \(h = 6\text{ cm}\).

Step 1: Expressing the slant height \(l\) in terms of height \(h\).
From our primary geometric properties, we know \(r = \frac{h}{3}\). Let us substitute this into the definition of slant height \(l\): \[ l = \sqrt{\left(\frac{h}{3}\right)^2 + h^2} = \sqrt{\frac{h^2}{9} + h^2} = \sqrt{\frac{10h^2}{9}} = \frac{\sqrt{10}}{3}h \]

Step 2: Expressing the curved surface area purely as a function of \(h\).
Now, substitute both \(r = \frac{h}{3}\) and \(l = \frac{\sqrt{10}}{3}h\) into the lateral surface area formula: \[ A_c = \pi \cdot \left(\frac{h}{3}\right) \cdot \left(\frac{\sqrt{10}}{3}h\right) \] Combine the constants and variables systematically: \[ A_c = \frac{\sqrt{10}\pi}{9} h^2 \]

Step 3: Differentiating with respect to time \(t\) and calculating the rate.
Differentiating implicitly with respect to time variable \(t\): \[ \frac{dA_c}{dt} = \frac{\sqrt{10}\pi}{9} \cdot \left( 2h \cdot \frac{dh}{dt} \right) \] Now plug in our set of known values at this exact instant (\(h = 6\text{ cm}\) and \(\frac{dh}{dt} = \frac{1}{40\pi}\text{ cm/s}\)): \[ \frac{dA_c}{dt} = \frac{\sqrt{10}\pi}{9} \cdot 2(6) \cdot \left(\frac{1}{40\pi}\right) \] Simplify the numbers: \[ \frac{dA_c}{dt} = \frac{12\sqrt{10}\pi}{9 \cdot 40\pi} \] Cancel out \(\pi\) from both the numerator and denominator: \[ \frac{dA_c}{dt} = \frac{12\sqrt{10}}{360} \] Reduce the fraction by dividing both the numerator and denominator by 12: \[ \frac{12}{12} = 1, \quad \frac{360}{12} = 30 \quad \Rightarrow \quad \frac{dA_c}{dt} = \frac{\sqrt{10}}{30}\text{ cm}^2\text{/s} \] This perfectly matches Option (A).