The circumference of a sphere was measured to be 76 cm with a possible error of 0.2cm.
1. Use differentials to estimate the maximum error in the calculated surface area.
Please round the answer to the nearest tenth.
2. What is the relative error in the calculated surface area?
_____cm^2
Please round the answer to the nearest tenth.
_________%
3.Use differentials to estimate the maximum error in the calculated volume.
Please round the answer to the nearest tenth.
______cm^3
4.What is the relative error in the calculated volume? Please round the answer to the nearest tenth.
_____%
C= 2PI r
dC= 2PI dr
dr= dC/2PI
S= 4PI r^2
dS= 8PI r dr= 8PI r/2PI dC=2C/PI dC
check my thinking
Solve for dS given C, and dC
i'll try it out thanks
To estimate the maximum error in the calculated surface area, we can use differentials.
Given that the circumference of the sphere is measured to be 76 cm with a possible error of 0.2 cm, we use the formula C = 2πr to find the radius.
C = 2πr
76 = 2πr
r = 76 / (2π)
Now we differentiate the surface area formula S = 4πr^2 with respect to r:
dS = 8πr dr
Substituting dr = dC / (2π) from the differential of the circumference, we get:
dS = 8πr (dC / (2π))
dS = 4r dC
To estimate the maximum error in surface area, we multiply the radius by the maximum error in circumference (0.2 cm):
max error in surface area = 4r * max error in circumference
max error in surface area = 4 * (76 / (2π)) * 0.2
Calculating this expression will give us the maximum error in the calculated surface area. Round it to the nearest tenth.
For the relative error in the calculated surface area, we divide the maximum error in surface area by the actual surface area and multiply by 100:
relative error in surface area = (max error in surface area / calculated surface area) * 100
To estimate the maximum error in the calculated volume, we can also use differentials. The formula for the volume of a sphere is V = (4/3)πr^3. We differentiate it with respect to r:
dV = 4πr^2 dr
Substituting dr = dC / (2π) from the differential of the circumference, we get:
dV = 4πr^2 (dC / (2π))
dV = 2r^2 dC
To estimate the maximum error in volume, we multiply the square of the radius by the maximum error in circumference:
max error in volume = 2r^2 * max error in circumference
max error in volume = 2 * ((76 / (2π))^2) * 0.2
Calculating this expression will give us the maximum error in the calculated volume. Round it to the nearest tenth.
For the relative error in the calculated volume, we divide the maximum error in volume by the actual volume and multiply by 100:
relative error in volume = (max error in volume / calculated volume) * 100
Now you can calculate the numerical values for the maximum error and relative error in both the surface area and volume of the sphere.