How Does the Radius of a Balloon Change with a 19.4% Increase in Surface Area?

jackt
Messages
1
Reaction score
0
a spherical balloon expands when it is taken from the cold outdoors to the inside of a warm house. if its surface area increases 19.4%, by what percentage does the radius of the balloon change?


I've tried this several times but I am not sure how to go about beginning the work
 
Physics news on Phys.org
If I were doing this problem,
I would write down quantities with labels like:

for the volume of the ball before...
V_{before}=\frac{4\pi}{3}r_{before}^3
and, similarly, for "after".
They can be compared by taking the ratio:
\frac{ V_{after} }{V_{before} }.

How would you use this strategy with your particular problem?

What does "increases 19.4%" mean in terms of a ratio?
 
The increase in 19.4% means the difference between the new surface area and the original or we could write it as:

4(pi)R^2-4(pi)r^2 where R is the new(larger) radius and r the original radius

We also know that increase percentage is calculated on the original value which happens to be 4(pi)r^2.

Hence, {4(pi)R^2-4(pi)r^2}/4(pi)r^2 *100=19.4

Evaluate this and then find the relation only in terms of the ratio of the difference in radius to the original radius...
 
Last edited:
ron_jay said:
in terms of the difference in radius...

For this type of problem, it's more efficient to think in ratios of radii, rather than difference in radii.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top