I have been having problems with this one question for the past few hours. I have been trying to find the radius, but I am not sure that is the way of solving this question. How could I go about this question.
A spherical balloon is to be filled with water so that its surface area increases at a constant rate of 1cm2 cm/s. - Find the volume when the volume is increasing at 10cm3 cm/s
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On
Here is an alternate approach:
The surface area of a sphere is $$S = 4\pi r^2$$ Differentiating with respect to time yields $$\frac{dS}{dt} = 8\pi r~\frac{dr}{dt} \tag{1}$$ Solving equation 1 for $dr/dt$ yields $$\frac{dr}{dt} = \frac{1}{8\pi r}~\frac{dS}{dt} \tag{2}$$ The volume of a sphere is $$V = \frac{4}{3}\pi r^3$$ Differentiating with respect to time yields $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \tag{3}$$ Substituting for $dr/dt$ yields $$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{1}{8\pi r}~\frac{dS}{dt} = \frac{r}{2}~\frac{dS}{dt} \tag{4}$$ Solving equation 4 for $r$ yields $$r = 2~\frac{\frac{dV}{dt}}{\frac{dS}{dt}}$$ Substituting $10~\text{cm}^3/\text{s}$ for $dV/dt$ and $1~\text{cm}^2/\text{s}$ for $dS/dt$ yields $$r = \frac{2 \cdot 10~\frac{\text{cm}^3}{\text{s}}}{1~\frac{\text{cm}^2}{\text{s}}} = 20~\text{cm}$$ Hence, the volume pf the sphere when the volume is increasing at a rate of $10~\text{cm}^3/\text{s}$ is $$V = \frac{4}{3}\pi(20~\text{cm})^3 = \frac{32000\pi}{3}~\text{cm}^3$$
Assume the balloon starts at 0 cm radius. After t seconds the balloon's surface area is t, hence the radius is $\sqrt{\frac t{4\pi}}$ and the volume $$\frac{4\pi}3\left(\frac t{4\pi}\right)^{3/2}$$ We must find where the derivative of this (wrt time) is 10: $$\frac{4\pi}3\times\frac32\left(\frac t{4\pi}\right)^{1/2}\times\frac1{4\pi}=10$$ $$\frac12\sqrt\frac t{4\pi}=10$$ $$\frac t{4\pi}=400$$ $$t=1600\pi$$ Hence the volume: $$\frac{4\pi}3\left(\frac{1600\pi}{4\pi}\right)^{3/2}=\frac{32000\pi}{3}.$$ The calculations if the balloon had a non-zero surface area to start are similar, and I leave this as an exercise.