The circumference of a sphere was measured to be 82 cm with a possible error of 0.5 cm. A. Use differentials to estimate the maximum error in the calculated volume.
What is the relative error?
B. Use differentials to estimate the maximum error in the calculated volume.
What is the relative error?

A) Maximum error = 170.32 cm³

B)Relative error = 0.0575

Step-by-step explanation:

A) Formula for circumference is: C = 2πr

Differentiating with respect to r, we have;

dC/dr = 2π

r is small, so we can write;

ΔC/Δr = 2π

So, Δr = ΔC/2π

We are told that ΔC = 0.5.

Thus; Δr = 0.5/2π = 0.25/π

Now, formula for Volume of a sphere is;

V(r) = (4/3)πr³

Differentiating with respect to r, we have;

dV/dr = 4πr²

Again, r is small, so we can write;

ΔS/Δr = 4πr²

ΔV = 4πr² × Δr

Rewriting, we have;

ΔV = ((2πr)²/π) × Δr

Since C = 2πr, we now have;

ΔV = (C²/π)Δr

ΔV will be maximum when Δr is maximum

Thus, ΔV = (C²/π) × 0.25/π

C = 82 cm

Thus;

ΔV = (82²/π) × 0.25/π

ΔV = 170.32 cm³

B) Formula for relative error = ΔV/V

Relative error = 170.32/((4/3)πr³)

Relative error = 170.32/((4/3)C³/8π³)

Relative errror = 170.32/((4/3)82³/8π³)

Relative error = 170.32/2963.744

Relative error = 0.0575

First, write the equation of the line containing the points (2,-5) and (-3,2).

We can use 2 point form, or point-slope form.

Let's use point-slope form.

the slope m is , then use any of the points to write the equation. (ex, pick (2, -5))

y-(-5)=(-7/5)(x-2)

y+5=(-7/5)x+14/5

y= (-7/5)x+14/5 - 5 =(-7/5)x+14/5 - 25/5 =(-7/5)x-11/5

Thus, the lines are 

i) y=-ax+4      and  ii) y=(-7/5)x-11/5

the slopes are the coefficients of x: -a and (-7/5),

the product of the slopes of 2 perpendicular lines is -1, 

so 

(-a)(-7/5)=-1

7/5a=-1

a=-1/(7/5)=-5/7

-5/7