Cars on a certain roadway travel on a circular arc of radius r. In order not to rely on friction alone to over come the centrifugal force, the road is banked at an angle of magnitude 0 from the horizontal. The banking angle must satisfy the equation
rg tan 0 = 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 d0/dt.
= v2
0 is angle I'm not sure how to input the angle symbol
Well, we could call the angle "theta" instead. So, the equation can be rewritten as:
r * g * tan(theta) = v^2
Now, let's differentiate both sides of the equation with respect to time t:
d/dt (r * g * tan(theta)) = d/dt (v^2)
Using the chain rule, we get:
r * g * sec^2(theta) * d(theta)/dt = 2v * dv/dt
Now, we need to solve for the relationship between dv/dt and d(theta)/dt:
dv/dt = (r * g * sec^2(theta) / (2v)) * d(theta)/dt
So, the relationship between the related rates dv/dt and d(theta)/dt is:
dv/dt = (r * g * sec^2(theta) / (2v)) * d(theta)/dt
To find the relationship between the related rates dv/dt and d0/dt, we can differentiate both sides of the equation rg tan θ = v^2 with respect to time t.
Differentiating rg tan θ = v^2 with respect to t, we get:
r(dθ/dt) * g * sec^2 θ = 2v(dv/dt)
Rearranging the equation, we get:
r(dθ/dt) = (2v(dv/dt)) / (g * sec^2 θ)
Recall that sec^2 θ = 1 + tan^2 θ. So we can substitute this value into the equation:
r(dθ/dt) = (2v(dv/dt)) / (g * (1 + tan^2 θ))
Now, we need to find the relationship between dθ/dt and dv/dt. To do this, we can use the relationship between the velocity v and the angle θ:
v = r(dθ/dt)
Differentiating v = r(dθ/dt) with respect to t, we get:
dv/dt = (d^2θ/dt^2) * r
Substituting this back into the equation above, we have:
r(dθ/dt) = (2(d^2θ/dt^2)r^2) / (g * (1 + tan^2 θ))
Canceling out the common factors of r, we get:
dθ/dt = (2(d^2θ/dt^2)) / (g * (1 + tan^2 θ))
Therefore, the relationship between the related rates dv/dt and dθ/dt is:
dv/dt = (2(d^2θ/dt^2)) / (g * (1 + tan^2 θ))
To find the relationship between the related rates dv/dt and d0/dt, we can differentiate the equation rg tan θ = v^2 with respect to time t.
Differentiating both sides of the equation with respect to t, we get:
r(dθ/dt)g tan θ + rg(sec^2 θ)(dθ/dt) = 2v(dv/dt)
Now, let's solve this equation to find the relationship between dv/dt and dθ/dt.
First, let's simplify the equation:
r(dθ/dt)g tan θ + rg(sec^2 θ)(dθ/dt) = 2v(dv/dt)
r(dθ/dt)g tan θ + r(dθ/dt)g = 2v(dv/dt)
r(dθ/dt)(g tan θ + g) = 2v(dv/dt)
r(dθ/dt)(g(tan θ + 1)) = 2v(dv/dt)
Next, let's substitute the equation rg tan θ = v^2:
r(dθ/dt)(g(tan θ + 1)) = 2(dθ/dt)v^2
Now, cancel out the common factor (dθ/dt):
r(g(tan θ + 1)) = 2vv'
Finally, solve for the relationship between dv/dt and dθ/dt:
dv/dt = [r(g(tan θ + 1))] / (2v)
Or, rearranging the equation:
dv/dt = r(g(tan θ + 1))/(2v)
This gives us the relationship between the related rates dv/dt and dθ/dt.
v^2 = rg tan(θ)
2v dv/dt = rg sec^2(θ) dθ/dt