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A conical paper cup is 10cm10\text{cm} tall with a radius of 20cm20\text{cm}. The bottom of the cup is punctured so that the water level goes down at a rate of 3cm/sec3\text{cm/sec}. At what rate is the volume of water in the cup changing when the water level is 4cm4\text{cm}?

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Q. A conical paper cup is 10cm10\text{cm} tall with a radius of 20cm20\text{cm}. The bottom of the cup is punctured so that the water level goes down at a rate of 3cm/sec3\text{cm/sec}. At what rate is the volume of water in the cup changing when the water level is 4cm4\text{cm}?
  1. 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=(13)πr2hV = (\frac{1}{3})\pi r^2 h, where rr is the radius and hh 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.
  2. Set Up Proportion for Similar Triangles: Given that the cup is 10 cm10\,\text{cm} tall with a radius of 20 cm20\,\text{cm}, we can use similar triangles to find the relationship between the radius (r)(r) and the height (h)(h) of the water. When the water level is at 4 cm4\,\text{cm} (h=4)(h = 4), we can set up a proportion using the dimensions of the full cup (radius 20 cm20\,\text{cm}, height 10 cm10\,\text{cm}) and the current water level. The proportion is (r/4)=(20/10)(r/4) = (20/10).
  3. Calculate Radius at Water Level extit{44cm}: Solving the proportion (r/4)=(20/10)(r/4) = (20/10) gives us r=(4×20)/10=8cmr = (4 \times 20) / 10 = 8\text{cm}. So when the water level is at 4cm4\text{cm}, the radius of the water surface is 8cm8\text{cm}.
  4. Find Rate of Change of Volume: Now we need to find the rate of change of the volume with respect to time, which is dVdt\frac{dV}{dt}. We know that the height of the water level is decreasing at a rate of dhdt=−3 cm/sec\frac{dh}{dt} = -3\,\text{cm/sec} (negative because the height is decreasing).
  5. Apply Chain Rule from Calculus: To find dVdt\frac{dV}{dt}, we need to use the chain rule from calculus, since VV is a function of rr, and rr is a function of hh. The chain rule tells us that dVdt=dVdr⋅drdh⋅dhdt\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dh} \cdot \frac{dh}{dt}.
  6. Calculate dV/drdV/dr: First, we find dV/drdV/dr. Differentiating V=(1/3)πr2hV = (1/3)\pi r^2h with respect to rr gives us dV/dr=(2/3)πrhdV/dr = (2/3)\pi rh.
  7. Calculate drdh\frac{dr}{dh}: Next, we find drdh\frac{dr}{dh} from our proportion. Differentiating r=410hr = \frac{4}{10}h with respect to hh gives us drdh=410\frac{dr}{dh} = \frac{4}{10}.
  8. 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=(23)πrh×(410)×(−3)dV/dt = (\frac{2}{3})\pi rh \times (\frac{4}{10}) \times (-3), where h=4h = 4cm and r=8r = 8cm.
  9. Simplify the Expression: Plugging in the values, we get dVdt=(23)π×8×4×(410)×(−3)=(23)π×32×(410)×(−3)=(23)π×32×(25)×(−3).\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).
  10. Simplify the Expression: Plugging in the values, we get dVdt=(23)π×8×4×(410)×(−3)=(23)π×32×(410)×(−3)=(23)π×32×(25)×(−3)\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 dVdt=(23)π×64×(−35)=−128π5 cm3/sec\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}.

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