Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the tangential and normal components \(\left(a_{T}\right.\) and \(a_{N}\) ) of the acceleration vector at \(t\). Then evaluate at \(t=t_{1}\). \(x(t)=1+3 t, y(t)=2-6 t ; t_{1}=2\)

Step-by-Step Solution

Answer

Both tangential and normal components of acceleration are 0 at \(t = 2\).

Step 1: Find the velocity vector

To find the velocity vector, we differentiate the position vector components with respect to time \(t\). This gives us \(v(t) = \left(\frac{dx}{dt}, \frac{dy}{dt}\right)\). For \(x(t)=1+3t\) and \(y(t)=2-6t\), the derivatives are \(\frac{dx}{dt} = 3\) and \(\frac{dy}{dt} = -6\). Therefore, the velocity vector is \(v(t) = (3, -6)\).

Step 2: Find the acceleration vector

The acceleration vector is found by differentiating the velocity vector components with respect to time \(t\). Since \(v(t) = (3, -6)\), both derivatives are zero, resulting in the acceleration vector \(a(t) = (0, 0)\).

Step 3: Evaluate the velocity and acceleration at \(t_1 = 2\)

At \(t = t_1 = 2\), the velocity vector remains \(v(2) = (3, -6)\) and the acceleration vector remains \(a(2) = (0, 0)\) because they are both constant.

Step 4: Calculate the magnitude of the velocity vector

The magnitude of the velocity vector is found using \(|v(t)| = \sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\).

Step 5: Calculate the tangential component of acceleration

The tangential component of acceleration is given by \(a_T = \frac{d|v|}{dt}\). Since the velocity magnitude \(3\sqrt{5}\) is constant, its derivative is zero, so \(a_T = 0\).

Step 6: Calculate the normal component of acceleration

The normal component of acceleration is given by \(a_N = \sqrt{|a(t)|^2 - a_T^2}\). Since \(a(t) = (0,0)\), we have \(|a(t)| = 0\) leading to \(a_N = \sqrt{0^2 - 0^2} = 0\).

Key Concepts

Velocity Vector

The velocity vector represents the rate of change of position in space over time. It's like capturing the direction and speed of an object as it moves. If you consider a path described by position functions such as \(x(t)\) and \(y(t)\), the velocity vector \(v(t)\) is found by differentiating these functions with respect to time \(t\).
For the given functions, \(x(t) = 1 + 3t\) and \(y(t) = 2 - 6t\), the differentiation yields \(\frac{dx}{dt} = 3\) and \(\frac{dy}{dt} = -6\).
Consequently, the velocity vector becomes \(v(t) = (3, -6)\).
The components of this vector, 3 and -6, tell us about the motion in the horizontal and vertical directions, respectively.

• The horizontal velocity is constant at 3 units per time.
• The vertical velocity is constant at -6 units per time, indicating downward motion.

Understanding the velocity vector is crucial before diving into other motion-related vectors or analyzing the motion further.

Acceleration Vector

The acceleration vector is a fundamental concept in understanding how velocity changes over time. This vector measures the rate of change of the velocity vector itself.
To find the acceleration vector, you differentiate the components of the velocity vector with respect to time \(t\).
Given that the velocity vector \(v(t) = (3, -6)\) holds constant values, the differentiation \(\frac{d}{dt} (3) = 0\) and \(\frac{d}{dt} (-6) = 0\) results in the acceleration vector \(a(t) = (0, 0)\).
Here are the highlights:

• The acceleration vector \((0,0)\) indicates no net change in velocity over time.
• In our case, this means steady-state motion without increase or decrease in speed in any direction.

This vector becomes especially important when analyzing more dynamic situations where velocities are changing.

Differentiation

Differentiation is a mathematical process you'll often use when working with vectors, such as velocity and acceleration.
In essence, it measures how a quantity changes over a small interval of time.
When you differentiate a function, you're essentially looking for its 'rate of change'. In the context of our exercise, you're applying differentiation one time to position functions to get velocity, and then again to those velocity functions to find acceleration.
Common steps include:

• For position functions \(x(t)\) and \(y(t)\), differentiate with respect to \(t\) to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), forming the velocity vector \((v_x, v_y)\).
• Differentiate these velocity components with respect to time again to get the acceleration vector.

Through differentiation, you systematically reveal layers of an object's motion, helping to understand both its current state and how it might evolve.

Magnitude of Velocity

The magnitude of a velocity vector essentially tells you the speed of the object. It is derived from the Pythagorean theorem and represents the overall speed without regard to direction.
For a velocity vector \(v(t) = (v_x, v_y)\), the magnitude is calculated as:
\[ |v(t)| = \sqrt{v_x^2 + v_y^2} \]
In our example, with \(v(t) = (3, -6)\), the magnitude is \(\sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\).
Key points about the magnitude of velocity:

• It provides a scalar value, the speed, representing how fast an object is moving.
• A constant magnitude, like in our example, indicates steady speed over time.

Understanding this concept can help immensely when you need to assess how fast a point on a curve is traveling, separate from its direction of travel.