A conical paper cup is $10\text{cm}$ tall with a radius of $20\text{cm}$ . The bottom of the cup is punctured so that the water level goes down at a rate of $3\text{cm/sec}$ . At what rate is the volume of water in the cup changing when the water level is $4\text{cm}$ ?

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Grade 6

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Q. A conical paper cup is

10\text{cm}

tall with a radius of

20\text{cm}

. The bottom of the cup is punctured so that the water level goes down at a rate of

3\text{cm/sec}

. At what rate is the volume of water in the cup changing when the water level is

4\text{cm}

Find Relationship Between Radius and Height: We need to find the rate of change of the volume of water in the cup, which is a conical shape. The formula for the volume of a cone is $V = (\frac{1}{3})\pi r^2 h$ , where $r$ is the radius and $h$ is the height. Since the radius changes as the water level changes, we need to find a relationship between the radius and the height of the water in the cone.
Set Up Proportion for Similar Triangles: Given that the cup is $10\,\text{cm}$ tall with a radius of $20\,\text{cm}$ , we can use similar triangles to find the relationship between the radius $(r)$ and the height $(h)$ of the water. When the water level is at $4\,\text{cm}$ $(h = 4)$ , we can set up a proportion using the dimensions of the full cup (radius $20\,\text{cm}$ , height $10\,\text{cm}$ ) and the current water level. The proportion is $(r/4) = (20/10)$ .
Calculate Radius at Water Level extit{ $4$ cm}: Solving the proportion $(r/4) = (20/10)$ gives us $r = (4 \times 20) / 10 = 8\text{cm}$ . So when the water level is at $4\text{cm}$ , the radius of the water surface is $8\text{cm}$ .
Find Rate of Change of Volume: Now we need to find the rate of change of the volume with respect to time, which is $\frac{dV}{dt}$ . We know that the height of the water level is decreasing at a rate of $\frac{dh}{dt} = -3\,\text{cm/sec}$ (negative because the height is decreasing).
Apply Chain Rule from Calculus: To find $\frac{dV}{dt}$ , we need to use the chain rule from calculus, since $V$ is a function of $r$ , and $r$ is a function of $h$ . The chain rule tells us that $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dh} \cdot \frac{dh}{dt}$ .
Calculate $dV/dr$ : First, we find $dV/dr$ . Differentiating $V = (1/3)\pi r^2h$ with respect to $r$ gives us $dV/dr = (2/3)\pi rh$ .
Calculate $\frac{dr}{dh}$ : Next, we find $\frac{dr}{dh}$ from our proportion. Differentiating $r = \frac{4}{10}h$ with respect to $h$ gives us $\frac{dr}{dh} = \frac{4}{10}$ .
Plug in Values into Chain Rule Equation: Now we can plug in the values we know into the chain rule equation. We have $dV/dt = (\frac{2}{3})\pi rh \times (\frac{4}{10}) \times (-3)$ , where $h = 4$ cm and $r = 8$ cm.
Simplify the Expression: Plugging in the values, we get $\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi \times 8 \times 4 \times \left(\frac{4}{10}\right) \times (-3) = \left(\frac{2}{3}\right)\pi \times 32 \times \left(\frac{4}{10}\right) \times (-3) = \left(\frac{2}{3}\right)\pi \times 32 \times \left(\frac{2}{5}\right) \times (-3).$
Simplify the Expression: Plugging in the values, we get $\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi \times 8 \times 4 \times \left(\frac{4}{10}\right) \times (-3) = \left(\frac{2}{3}\right)\pi \times 32 \times \left(\frac{4}{10}\right) \times (-3) = \left(\frac{2}{3}\right)\pi \times 32 \times \left(\frac{2}{5}\right) \times (-3)$ . Simplifying the expression, we get $\frac{dV}{dt} = \left(\frac{2}{3}\right)\pi \times 64 \times \left(-\frac{3}{5}\right) = -\frac{128\pi}{5} \text{ cm}^3/\text{sec}$ .