Comprehension

At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

Question: 1

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 \(\alpha\).

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In problems involving conical containers, the ratio of radius to height remains completely invariant at all water levels: \[ \frac{r}{h} = \frac{\text{Full Radius}}{\text{Full Height}} \] This allows you to quickly substitute out variables in calculus rate-of-change equations without reworking the geometry every time.

Updated On: Jul 8, 2026

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Question: 2

At what rate is the juice level in the cup rising when the juice is 6 cm deep?

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Always convert decimal rate inputs like \(0.1\) to fractions (\(\frac{1}{10}\)) right away. Fractions are far easier to manipulate, reduce, and cross-multiply when solving calculus tracking equations.

Updated On: Jul 8, 2026

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Question: 3

When the juice is 6 cm deep, then find at what rate is the upper surface area of juice increasing?

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Differentiating with respect to the radius directly via \(\frac{dA_s}{dt} = 2\pi r \frac{dr}{dt}\) can sometimes be less error-prone than maintaining fractional squares of \(h\), provided you compute \(r\) and \(\frac{dr}{dt}\) accurately beforehand.

Updated On: Jul 8, 2026

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Question: 4

When the juice is 6 cm deep, then find the rate at which the wetted surface area of the cup is increasing.

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The slant height of a cone always retains a fixed proportional ratio to its vertical height, just like the radius does. Finding \(l = k \cdot h\) early eliminates complex square root derivatives later on.

Updated On: Jul 8, 2026

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