I am having difficulty with this math question in related rates. can anyone help?

Question

The combined electrical resistance R of R1 and R2, connected in parallel, is
1/R= 1/R1 + 1/R2
where R, R1 and R2 are measured in ohms. R1 and R2 are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is R changing when R1= 50 ohms and R2= 75 ohms?

Answer 1

The formula you're given is:

1/R= 1/R1 + 1/R2

So you need to know what R is when R1 = 50Ω and R2 = 75Ω:

1/R = 1/50Ω + 1/75Ω = 0.03333 S

Therefore:

R = 30Ω

For implicit differentiation, I think it's easier to write the original equation as:

Rˉ¹ = R1ˉ¹ + R2ˉ¹

Then, when you differentiate, you get:

-Rˉ²(R') = - R1ˉ²(R1') - R2ˉ²(R2')

Dividing through by -1 to get rid of all the negatives:

Rˉ²(R') = R1ˉ²(R1') + R2ˉ²(R2')

Now substitute R = 30Ω, R1 = 50Ω, R2 = 75Ω, R1' = 1Ω/s, R2' = 1.5Ω/s:

30ˉ²(R') = 50ˉ²(1) + 75ˉ²(1.5)

Multiply both sides by 30², and plug the numbers into a calculator:

R' = 30²(1/50² + 1.5/75²)
= 30²/50² + 1.5(30²/75²)
= 9/25 + 3/2(4/25)
= 9/25 + 6/25
= 15/25
= 0.6 Ω/s

Answer 2

Given

1/R = 1/R1 + 1/R2
dR1/dt = 1 Ω/sec
dR2/dt = 1.5 Ω/sec

Find dR/dt
when R1 = 50Ω and R2 = 75Ω

1/R = 1/R1 + 1/R2
Solve for R.
(R1)(R2) = R(R2) + R(R1) = R(R1 + R2)
R = (R1)(R2)/(R1 + R2)
R = (50)(75)/(50 + 75) = 30

Differentiating implicitly we have:

(-1/R²)(dR/dt) = (-1/R1²)(dR1/dt) - (1/R2²)(dR2/dt)
dR/dt = (R²/R1²)(dR1/dt) + (R²/R2²)(dR2/dt)
dR/dt = (R²/R1²)(dR1/dt) + (R²/R2²)(dR2/dt)
dR/dt = (30²/50²)(1) + (30²/75²)(1.5)
dR/dt = 9/25 + (4/25)(3/2) = 9/25 + 6/25
= 15/25 = 3/5 = 0.60 Ω/sec

Answer 3

Just as R = (R1*R2)/(R1+R2) = 30 ohms so also do the rate of changes vary so that:
dR = (dR1*dR2)/(dR1+dR2)
dR = 1*1.5/(1+1.5) = 1.5/2.5 = 0.6 ohms/sec