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

Question

Cars on a certain roadway travel on a circular arc of radius r. In order not to rely on friction alone to overcome the centrifugal force, the road is banked at an angle of magnitude Q (theta) from the horizontal. The banking angle must satisfy the equation
rg tanQ= v^2
where v is the velocity of the cars and g= 32 ft/sec^2 is the acceleration due to gravity. Find the relationship between the related rates dv/dt and dQ/dt.

Answer 1

Well, r and g don't change with respect to time, so this one is pretty easy.

32r tan(Q) = v²

Differentiating implicitly:

32r sec²(Q) (Q') = 2v(v')

So, solving for v' in terms of Q':

v' = Q' (16r sec²(Q)/v)

Answer 2

Differentiate implicitly.

rg tan Q= v²
rg(sec² Q)(dQ/dt) = 2v(dv/dt)
dQ/dt = 2v(dv/dt)/{rg(sec² Q)}
(dQ/dt)/(dv/dt) = 2v/{rg(sec² Q)}

Plugging in the value for g we get

(dQ/dt)/(dv/dt) = 2v/{32r(sec² Q)}
(dQ/dt)/(dv/dt) = v/{16r(sec² Q)}

Answer 3

Given a million/R = a million/R1 + a million/R2 dR1/dt = a million ?/sec dR2/dt = a million.5 ?/sec locate dR/dt while R1 = 50? and R2 = 75? a million/R = a million/R1 + a million/R2 sparkling up 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've: (-a million/R²)(dR/dt) = (-a million/R1²)(dR1/dt) - (a million/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)(a million) + (30²/75²)(a million.5) dR/dt = 9/25 + (4/25)(3/2) = 9/25 + 6/25 = 15/25 = 3/5 = 0.60 ?/sec