Question
(a) Topic-8.2-Connecting Position Velocity and Acceleration of Functions Using Integrals
(b) Topic-9.1-Defining and Differentiating Parametric Equations
(c) Topic-9.6-Solving Motion Problems Using Parametric and Vector-Valued Functions
(d) Topic-9.3-Finding Arc Lengths of Curves Given by Parametric Equations
The graph of \( y \), consisting of three line segments, is shown in the figure above. \\
At \( t = 0 \), the particle is at position \( (5, 1) \).
(a) Find the position of the particle at \( t = 3 \).
(b) Find the slope of the line tangent to the path of the particle at \( t = 3 \).
(c) Find the speed of the particle at \( t = 3 \).
(d) Find the total distance traveled by the particle from \( t = 0 \) to \( t = 2 \).
▶️Answer/Explanation
\(\textbf{2(a)}\)
The position of the particle at \( t = 3 \) is \( (x(3), y(3)) \).
\(
x(3) = x(0) + \int_0^3 x'(t) \, dt \approx 5 + 9.377035 \approx 14.37703509
\)
Since the slope of the segment connecting \( (2, -1) \) and \( (4, 0) \) is \( \frac{1}{2} \), then \( y(3) = -\frac{1}{2} \).
So the position of the particle at \( t = 3 \) is
\(
\boxed{\left( 14.377, -\frac{1}{2} \right)}.
\)
\(\textbf{(b)}\)
\(
y'(3) = \frac{1}{2} \quad \text{from part (a)}.
\)
\(
x'(3) \approx 9.956375928 \quad \text{using the given expression for } \frac{dx}{dt}.
\)
\(
m_{t=3} = \frac{dy}{dx} \bigg|_{t=3} = \frac{y'(3)}{x'(3)} \approx \frac{\frac{1}{2}}{9.956375928} = \boxed{0.0502190761}.
\)
\(\textbf{(c)}\)
\(
\text{Speed}_{t=3} = \sqrt{(x'(3))^2 + (y'(3))^2} \approx \sqrt{(9.956375928)^2 + \left(\frac{1}{2}\right)^2} \approx \boxed{9.968922791}.
\)
\(\textbf{(d)}\)
\(
\text{Total Distance} = \int_0^2 \sqrt{(x'(t))^2 + (y'(t))^2} \, dt = \int_0^1 \sqrt{(x'(t))^2 + (-2)^2} \, dt + \int_1^2 \sqrt{(x'(t))^2 + 0^2} \, dt \approx \boxed{4.34987}.
\)