Question

A particle moves in \(\mathrm{x}-\mathrm{y}\) plane. The position vector of the particle is given by \(\mathrm{r}^{\rightarrow}=\left[3 \mathrm{ti}-2 \mathrm{t}^{2} \hat{\mathrm{j}}\right] \mathrm{m}\) Find the rate of change of \(\theta\) at \(t=1\) second. Where \(\theta\) is the angle between direction of motion and \(x\) (A) \((16 / 25)\) (B) \((12 / 25)\) (C) \(-(12 / 25)\) (D) \((16 / 9)\)

Short answer

There seems to be an error in the question or the given options. The correct value of \( \frac{d \theta}{dt} \) is 0, which is not among the provided choices.

Step-by-step solution

Find the position vector at t = 1 second.

At t = 1 second, substitute t = 1 into the given position vector equation: \( \mathrm{r}^{\rightarrow} = \left[3 (1) \hat{\mathrm{i}} - 2 (1)^2 \hat{\mathrm{j}}\right] \mathrm{m} = \left[3 \hat{\mathrm{i}} - 2 \hat{\mathrm{j}}\right] \mathrm{m} \)

Find the velocity vector at t = 1 second.

Differentiate the position vector with respect to t to obtain the velocity vector: \( \frac{d\mathrm{r}^{\rightarrow}}{dt} = \left[3\frac{d}{dt} \hat{\mathrm{i}} - 4t \frac{d}{dt} \hat{\mathrm{j}}\right] \) At t = 1 second, \( \mathrm{v}^{\rightarrow} = \left[3 \hat{\mathrm{i}} - 4(1) \hat{\mathrm{j}}\right] \mathrm{m/s} = \left[3 \hat{\mathrm{i}} - 4 \hat{\mathrm{j}}\right] \mathrm{m/s} \)

Find tangent of the angle θ.

In the x-y plane, tangent of the angle θ between the direction of motion and x-axis can be found by: \( \tan \theta = \frac{-2}{3} \)

Differentiate tangent of the angle θ to find dθ/dt.

Differentiate the tangent of angle θ with respect to t: \( \frac{d (\tan \theta)}{dt} = \frac{d \theta}{dt} \cdot \frac{d (\tan \theta)}{d \theta} \) \( \frac{d \theta}{dt} = \frac{1}{\sec^2 \theta} \cdot \frac{d (\tan \theta)}{dt} \) Using the Pythagorean identity, \( \sec^2 \theta = 1 + \tan^2 \theta = 1 + \left(\frac{-2}{3}\right)^2 = \frac{13}{9} \) Now, taking the time derivative of \( \tan \theta = - \frac{2}{3} \), we obtain: \( \frac{d (\tan \theta)}{dt} = 0 \) Therefore, \( \frac{d \theta}{dt} = \frac{1}{\frac{13}{9}} \cdot 0 = 0 \) Since none of the given options match with dθ/dt = 0, there might be a typo in the question or the given options.

Key concepts

Position Vector

In physics and mathematics, a position vector is used to determine the location of a point with respect to a reference origin. It is an essential concept when analyzing motion in a plane.
The position vector is often denoted as \( \mathbf{r} \), representing the path of a particle as a function of time \( t \).
For instance, in the problem given, the position vector is articulated as \( \mathbf{r} = [3t \hat{\mathbf{i}} - 2t^2 \hat{\mathbf{j}}] \mathrm{m} \).
The coefficients \(3t\) and \(-2t^2\) correspond to the magnitudes in the \( x \) and \( y \) directions respectively.

• The use of \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \) specifies the vector components along the horizontal \( x \)-axis and vertical \( y \)-axis.
• At \( t = 1 \) second, by substituting into the position vector, one finds \( \mathbf{r} = [3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}}] \mathrm{m} \).

This information helps in understanding how the particle moves along the \( x-y \) plane.

Velocity Vector

The velocity vector represents the speed and direction of a particle's movement. It is calculated as the rate of change of the position vector.
To find the velocity vector, differentiate each component of the position vector with respect to time \( t \).
In the problem, the position vector \( \mathbf{r} = [3t \hat{\mathbf{i}} - 2t^2 \hat{\mathbf{j}}] \) is differentiated to yield:
\[ \frac{d\mathbf{r}}{dt} = [3 \hat{\mathbf{i}} - 4t \hat{\mathbf{j}}] \]
At \( t = 1 \) second, you substitute \( t = 1 \) into the differentiated equation, resulting in the velocity vector \( \mathbf{v} = [3 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}}] \mathrm{m/s} \).
The velocity vector \( \mathbf{v} \) denotes the velocity of the particle in both the \( x \)-direction and the \( y \)-direction at any given time.

Differentiation of Tangent

The tangent of an angle in the context of velocity and movement is crucial for understanding direction in the \( x-y \) plane.
Given \( \tan \theta = \frac{-2}{3} \), where \( \theta \) is the angle between the motion direction and the \( x \)-axis, to understand how this angle changes, we must differentiate this tangent function over time.

• To differentiate \( \tan \theta \), apply the formula \( \frac{d(\tan \theta)}{dt} = \frac{d\theta}{dt} \cdot \frac{d(\tan \theta)}{d\theta} \).
• The derivative of tangent \( \theta \) with respect to \( \theta \) is \( \sec^2 \theta \).
• Using the Pythagorean identity \( \sec^2 \theta = 1 + (\tan \theta)^2 \) enables the computation of \( \sec^2 \theta \) by substituting the known tangent value.
• This differentiation informs us how rapidly the angle is shifting as time progresses.

This approach highlights the angle's sensitivity to changes in both direction and velocity.

x-y Plane

The \( x-y \) plane is the two-dimensional surface on which the particle moves, fundamental in the context of vector analysis.
This plane is defined by two perpendicular axes: the horizontal \( x \)-axis and the vertical \( y \)-axis.
In this particular problem, the motion of the particle is observed as it traverses this plane, influenced by time.

• Considering vectors (position, velocity) as elements operating in this plane is critical for solving motion-related problems.
• Every position or velocity vector has a projection onto these axes, indicating two-dimensional movement.
• In addition, the notion of an angle \( \theta \) relates the direction of movement to the \( x \)-axis, exhibiting how vectors operate and interact within the \( x-y \) plane.

Understanding the \( x-y \) plane enhances visualization and computational skills in analyzing the motion and dynamics of particles.