Respuesta :
The magnitude of the volume of the balloon increases at a faster rate than
the radius of the balloon.
- The rate of change of the volume of the balloon with respect to radius is not constant.
Reasons:
The rate at which the volume of the balloon is increasing, [tex]\dfrac{dV}{dt} = 15 \ cm^3/s[/tex]
The initial volume of the balloon, V = 30 cm³
The rate of change of the volume of the balloon with radius is given as follows;
[tex]\dfrac{\Delta V}{\Delta r}= \dfrac{V_2 - V_1}{r_2 - r_1}[/tex]
[tex]Volume \ of \ the \ balloon, \ V = \dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
[tex]V = 30 + \dfrac{dV}{dt} \times t[/tex]
∴ V = 30 + 15·t
At t = 0, V = 30 cm³
Where;
[tex]r =\sqrt[3]{ \dfrac{3 \cdot V}{4 \cdot \pi} }[/tex]
When V = 30
- [tex]r =\sqrt[3]{ \dfrac{3 \times 30}{4 \cdot \pi} } \approx 1.93[/tex]
The radius, r ≈ 1.93 cm.
At t = 10 s., we have;
V = 30 + 15 × 10 = 180
- [tex]r =\sqrt[3]{ \dfrac{3 \times 180}{4 \cdot \pi} } \approx 3.5[/tex]
At t = 20 s., we have;
V = 30 + 15 × 20 = 330
- [tex]r =\sqrt[3]{ \dfrac{3 \times 330}{4 \cdot \pi} } \approx 4.29[/tex]
Therefore;
Between 0 s. and 10 s., we have;
- [tex]\dfrac{\Delta V}{\Delta r}=\dfrac{180 - 30}{3.5-1.93} \approx 96[/tex]
Rate of change of the volume of the balloon with respect to the radius of the balloon to the nearest centimeter, between 0 s and 10 s is 96 cm³/cm.
Between 10 s. and 20 s., we have;
- [tex]\dfrac{\Delta V}{\Delta r}=\dfrac{330 - 180}{4.29 - 3.5} \approx 190[/tex]
Between 10 s. and 20 s. the rate of change is approximately 190 cm³/cm.
Therefore;
The rate of change of the volume of the balloon with respect to radius is not constant, but increases as the volume of the balloon increases.
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