Question

\( \text { Radway Design } \) 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 \(\theta\) from the horizontal (see figure). The banking angle must satisfy the equation \(r g \tan \theta=v^{2},\) where \(v\) is the velocity of the cars and \(g=32\) feet per second per second is the acceleration due to gravity. Find the relationship between the related rates \(d v / d t\) and \(d \theta / d t\)

Short answer

The relationship between the rates of change of \(v\) and \(\theta\) is \(\frac{dv}{dt} = \frac{2v}{\sec^2 \theta} \frac{d\theta}{dt}\)

Step-by-step solution

Rewrite the given equation

The given equation is \(rg \tan \theta = v^2\). It can be rewritten as \(v = \sqrt{rg \tan \theta}\).

Differentiate the equation with respect to \(t\)

Differentiating both sides of the equation with respect to \(t\) will give us \(\frac{dv}{dt} = \frac{1}{2\sqrt{rg \tan \theta}} * [rg \sec^2 \theta * \frac{d\theta}{dt}]\). In this step, the chain rule was used to differentiate \(\sqrt{rg \tan \theta}\).

Simplify the equation

Simplify \(\frac{dv}{dt} = \frac{1}{2\sqrt{rg \tan \theta}} * [rg \sec^2 \theta * \frac{d\theta}{dt}]\) to \(\frac{dv}{dt} = \frac{1}{2v} \sec^2 \theta \frac{d\theta}{dt}\).

Find the relationship between \(dv/dt\) and \(d\theta/dt\)

To get the relationship between \(dv/dt\) and \(d\theta/dt\), rearrange \(\frac{dv}{dt} = \frac{1}{2v} \sec^2 \theta \frac{d\theta}{dt}\) and we get \(\frac{dv}{dt} = \frac{2v}{\sec^2 \theta} \frac{d\theta}{dt}\).