Problem Overview
We evaluate the required angular velocity d θ d t \frac{d\theta}{dt} d t d θ
θ ( t ) = tan − 1 ( x ( t ) ) , x ( t ) = 2 t t + 1 \theta(t) = \tan^{-1}(x(t)), \qquad x(t) = \frac{2t}{t+1} θ ( t ) = tan − 1 ( x ( t )) , x ( t ) = t + 1 2 t Step-by-Step Mathematical Solution
Step 1: Establish Chain Rule Blueprint
The pivot angle θ \theta θ x x x t t t
d θ d t = d θ d x ⋅ d x d t \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt} d t d θ = d x d θ ⋅ d t d x Step 2: Differentiate Outer Inverse Tangent Layer
Using the standard derivative rule for arctangent:
d θ d x = 1 1 + x 2 \frac{d\theta}{dx} = \frac{1}{1 + x^2} d x d θ = 1 + x 2 1 Step 3: Differentiate Inner Position Function via Quotient Rule
The function x ( t ) = 2 t t + 1 x(t) = \frac{2t}{t+1} x ( t ) = t + 1 2 t u = 2 t u = 2t u = 2 t v = t + 1 v = t+1 v = t + 1
u ′ = 2 u' = 2 u ′ = 2 v ′ = 1 v' = 1 v ′ = 1
Applying the Quotient Rule u ′ v − u v ′ v 2 \frac{u'v - uv'}{v^2} v 2 u ′ v − u v ′
d x d t = ( 2 ) ( t + 1 ) − ( 2 t ) ( 1 ) ( t + 1 ) 2 = 2 t + 2 − 2 t ( t + 1 ) 2 = 2 ( t + 1 ) 2 m/s \frac{dx}{dt} = \frac{(2)(t+1) - (2t)(1)}{(t+1)^2} = \frac{2t + 2 - 2t}{(t+1)^2} = \frac{2}{(t+1)^2}\text{ m/s} d t d x = ( t + 1 ) 2 ( 2 ) ( t + 1 ) − ( 2 t ) ( 1 ) = ( t + 1 ) 2 2 t + 2 − 2 t = ( t + 1 ) 2 2 m/s Step 4: Layer Components and Evaluate at t = 1 second t = 1\text{ second} t = 1 second
Find item position x ( 1 ) x(1) x ( 1 ) t = 1 t = 1 t = 1
x ( 1 ) = 2 ( 1 ) 1 + 1 = 2 2 = 1 meter x(1) = \frac{2(1)}{1 + 1} = \frac{2}{2} = 1\text{ meter} x ( 1 ) = 1 + 1 2 ( 1 ) = 2 2 = 1 meter Evaluate d θ d x \frac{d\theta}{dx} d x d θ x = 1 x = 1 x = 1
d θ d x ∣ x = 1 = 1 1 + ( 1 ) 2 = 1 2 rad/m \frac{d\theta}{dx}\Big|_{x=1} = \frac{1}{1 + (1)^2} = \frac{1}{2}\text{ rad/m} d x d θ x = 1 = 1 + ( 1 ) 2 1 = 2 1 rad/m Evaluate d x d t \frac{dx}{dt} d t d x t = 1 t = 1 t = 1
d x d t ∣ t = 1 = 2 ( 1 + 1 ) 2 = 2 4 = 1 2 m/s \frac{dx}{dt}\Big|_{t=1} = \frac{2}{(1+1)^2} = \frac{2}{4} = \frac{1}{2}\text{ m/s} d t d x t = 1 = ( 1 + 1 ) 2 2 = 4 2 = 2 1 m/s Compute instantaneous angular tracking velocity d θ d t \frac{d\theta}{dt} d t d θ
d θ d t = d θ d x ⋅ d x d t = ( 1 2 ) ⋅ ( 1 2 ) = 1 4 = 0.25 rad/s \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt} = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{4} = 0.25\text{ rad/s} d t d θ = d x d θ ⋅ d t d x = ( 2 1 ) ⋅ ( 2 1 ) = 4 1 = 0.25 rad/s
Engineering Conclusion
At t = 1 second t = 1\text{ second} t = 1 second x = 1 m x = 1\text{ m} x = 1 m d θ d t = 0.25 rad/s \frac{d\theta}{dt} = 0.25\text{ rad/s} d t d θ = 0.25 rad/s 14.32 ∘ /s 14.32^\circ\text{/s} 14.3 2 ∘ /s