If the radius of a sphere decreases by 0.1%, find the percentage decrease in the surface area and volume
Let's assume the original radius of the sphere is r.
The surface area of a sphere is given by the formula 4πr².
The volume of a sphere is given by the formula (4/3)πr³.
If the radius decreases by 0.1%, the new radius would be (r - 0.1/100 * r) = (r - 0.001r) = 0.999r.
1. Percentage Decrease in Surface Area:
New surface area = 4π(0.999r)² = 3.996πr²
% decrease in surface area = ((4πr² - 3.996πr²)/ (4πr²)) × 100%
= ((0.004πr²)/ (4πr²)) × 100%
= 0.1%
Therefore, the surface area of the sphere decreases by 0.1%.
2. Percentage Decrease in Volume:
New volume = (4/3)π(0.999r)³ = 0.999π(4/3)r³ = (0.999)(4/3)πr³
% decrease in volume = ((4/3)πr³ - (0.999)(4/3)πr³)/ ((4/3)πr³) × 100%
= ((4/3)πr³ - (3.996/3)πr³)/ ((4/3)πr³) × 100%
= (0.004/3) × 100%
≈ 0.1333%
Therefore, the volume of the sphere decreases by approximately 0.1333%.
To find the percentage decrease in the surface area and volume of a sphere when its radius decreases by 0.1%, we need to follow these steps:
Step 1: Understand the formulas for the surface area and volume of a sphere.
The surface area (A) of a sphere is given by the formula:
A = 4πr^2
The volume (V) of a sphere is given by the formula:
V = (4/3)πr^3
Step 2: Calculate the original surface area and volume.
Let's assume the original radius of the sphere is "r".
Original surface area (A1) = 4πr^2
Original volume (V1) = (4/3)πr^3
Step 3: Calculate the new surface area and volume.
Since the radius decreases by 0.1%, the new radius (r2) is equal to 99.9% of the original radius (r).
New surface area (A2) = 4πr2^2 = 4π(0.999r)^2 = 4π(0.998001)r^2 = 0.998001 * 4πr^2 = 0.998001 * A1
New volume (V2) = (4/3)πr2^3 = (4/3)π(0.999r)^3 = (4/3)π(0.997003001)r^3 = 0.997003001 * (4/3)πr^3 = 0.997003001 * V1
Step 4: Calculate the percentage decrease.
Percentage decrease in surface area = ((A1 - A2) / A1) * 100
Percentage decrease in volume = ((V1 - V2) / V1) * 100
Now, we can substitute the values and calculate the percentage decrease in surface area and volume.