if the radius of a sphere is increased by 10%, prove that its volume will be increased by33.1%.
V=4/3 ¦Ðr^3
r -> 1.1 r
V=4/3 ¦Ð¡¼(1.1r)¡½^3
V=4/3 ¦Ð x 1.331 r^3
So the increase is .331 or 33.1 %
How it will come 1.1 or 11/10
To prove that the volume of a sphere increases by 33.1% when the radius is increased by 10%, we can use the formulas for the volume of a sphere and the relationship between the radius and volume.
The formula for the volume of a sphere is:
V = (4/3)πr^3
Where V is the volume and r is the radius.
Let's assume the initial radius of the sphere is r.
If the radius is increased by 10%, the new radius would be r + 0.1r = 1.1r.
To find the new volume of the sphere, we substitute the new radius into the volume formula:
V_new = (4/3)π(1.1r)^3
= (4/3)π(1.1^3)(r^3)
= (4/3)π(1.331)(r^3)
= (4/3)π(1.331r^3)
Now, let's calculate the percentage increase in volume:
Increase in volume = V_new - V
= (4/3)π(1.331r^3) - (4/3)πr^3
= (4/3)π(1.331r^3 - r^3)
= (4/3)π(0.331r^3)
Percentage increase in volume = (Increase in volume / V) * 100%
= [(4/3)π(0.331r^3) / (4/3)πr^3] * 100%
= [(0.331r^3) / (r^3)] * 100%
= 0.331 * 100%
= 33.1%
Therefore, when the radius of a sphere is increased by 10%, its volume will be increased by approximately 33.1%.
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