Math Problem Statement

A balloon is in the form of a right circular cylinder of radius 1.5 m, length 4 m, and is surrounded by hemispherical ends. If the radius is increased by 0.01 m and length by 0.05 m, find the percentage of change of volume by balloon.

Solution

The problem describes a balloon in the shape of a right circular cylinder with hemispherical ends, with an initial radius of r=1.5mr = 1.5 \, \text{m} and a length (excluding the hemispheres) of h=4mh = 4 \, \text{m}. The radius and length increase by 0.01m0.01 \, \text{m} and 0.05m0.05 \, \text{m}, respectively. The task is to find the percentage change in the balloon's volume after these dimensional increases.

Step-by-Step Solution

  1. Volume Calculation for Original Shape:

    • The balloon is made up of a cylindrical part and two hemispheres.
    • Volume of the cylinder: Vcyl=πr2hV_{\text{cyl}} = \pi r^2 h.
    • Volume of each hemisphere: Vhemisphere=23πr3V_{\text{hemisphere}} = \frac{2}{3} \pi r^3.
    • Since there are two hemispheres, the total volume for them is Vhemispheres=2×23πr3=43πr3V_{\text{hemispheres}} = 2 \times \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3.
    • Total initial volume VV of the balloon: V=Vcyl+Vhemispheres=πr2h+43πr3V = V_{\text{cyl}} + V_{\text{hemispheres}} = \pi r^2 h + \frac{4}{3} \pi r^3
  2. Differentiating Volume with Respect to Radius and Height:

    • Let VV be a function of rr and hh: V(r,h)=πr2h+43πr3V(r, h) = \pi r^2 h + \frac{4}{3} \pi r^3.
    • To find the approximate change in volume (ΔV\Delta V), we can use the multivariable differentiation formula: ΔVVrΔr+VhΔh\Delta V \approx \frac{\partial V}{\partial r} \Delta r + \frac{\partial V}{\partial h} \Delta h
    • Compute Vr\frac{\partial V}{\partial r} and Vh\frac{\partial V}{\partial h}:
      • Vr=2πrh+4πr2\frac{\partial V}{\partial r} = 2 \pi r h + 4 \pi r^2
      • Vh=πr2\frac{\partial V}{\partial h} = \pi r^2
  3. Substitute Values:

    • Given r=1.5mr = 1.5 \, \text{m}, h=4mh = 4 \, \text{m}, Δr=0.01m\Delta r = 0.01 \, \text{m}, and Δh=0.05m\Delta h = 0.05 \, \text{m}.
    • Calculate ΔV\Delta V using the derived partial derivatives: ΔV(2π1.54+4π1.52)0.01+(π1.52)0.05\Delta V \approx (2 \pi \cdot 1.5 \cdot 4 + 4 \pi \cdot 1.5^2) \cdot 0.01 + (\pi \cdot 1.5^2) \cdot 0.05
  4. Compute the Percentage Change:

    • Initial volume VV: V=π(1.5)24+43π(1.5)3V = \pi \cdot (1.5)^2 \cdot 4 + \frac{4}{3} \pi \cdot (1.5)^3
    • Percentage change in volume: Percentage Change=(ΔVV)×100\text{Percentage Change} = \left( \frac{\Delta V}{V} \right) \times 100

By substituting the values and solving, we obtain the percentage change in the balloon’s volume.

Would you like to see each computation step in detail or have further clarification?

Additional Questions to Explore

  1. How would the percentage change in volume differ if only the radius increased?
  2. What if the balloon were made of a different geometric combination, like a cone with hemispherical ends?
  3. How does the change in volume depend on the rate of increase in radius versus height?
  4. What is the effect on volume if the radius change were in a negative direction (decreasing radius)?
  5. Can this method of approximation be used for larger changes in dimensions?

Tip:

For small changes in dimensions, using partial derivatives is a quick way to approximate the volume change without recalculating the exact volume after each dimension change.

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Math Problem Analysis

Mathematical Concepts

Geometry
Volume Calculation
Differentiation

Formulas

Volume of a cylinder: V_cyl = πr^2h
Volume of a hemisphere: V_hemisphere = (2/3)πr^3
Total volume of balloon: V = πr^2h + (4/3)πr^3
Approximate volume change: ΔV ≈ (∂V/∂r)Δr + (∂V/∂h)Δh

Theorems

Multivariable Differentiation

Suitable Grade Level

Grade 11-12