A particle is moving with velocity v = k(yi + xj), where k is a constant. The general equation for i

Q: Why do we divide $v_y$ by $v_x$?

We want to find the equation of the path, which is a relationship between $y$ and $x$ independent of time $t$. By dividing $v_y = \frac{dy}{dt}$ by $v_x = \frac{dx}{dt}$, the $dt$ terms cancel out, giving us the slope $\frac{dy}{dx}$ purely in terms of $x$ and $y$.

Q: What does the constant of integration represent physically?

The constant of integration is determined by the initial conditions of the particle. Specifically, it depends on the exact $(x, y)$ coordinates of the particle at time $t = 0$. Different starting points will yield different constants, placing the particle on different hyperbolic paths.

Q: What shape does the equation $y^2 = x^2 + c$ represent?

It represents a family of hyperbolas. If $c > 0$, the hyperbolas open upwards and downwards along the y-axis. If $c < 0$, they open leftwards and rightwards along the x-axis. If $c = 0$, the equation becomes $y^2 = x^2$, which represents two intersecting straight lines $y = x$ and $y = -x$.

Q: Can we solve this by finding $x(t)$ and $y(t)$ first?

Yes, but it's much more complicated. You would have to solve a system of coupled differential equations: $\frac{dx}{dt} = ky$ and $\frac{dy}{dt} = kx$. Taking the derivative of the first gives $\frac{d^2x}{dt^2} = k\frac{dy}{dt} = k^2x$. This is a second-order differential equation whose solutions involve exponential functions $e^{kt}$ and $e^{-kt}$. Eliminating $t$ from those solutions will eventually give you the same hyperbola, but finding $\frac{dy}{dx}$ directly is much faster.

Q: Why is the constant $2C'$ just written as 'constant' in the final step?

In mathematics, an arbitrary constant multiplied by a fixed number is still just an arbitrary constant. Since we don't know the value of $C'$ until we are given initial conditions, $2C'$ is equally unknown. We absorb the factor of 2 into the constant for a cleaner final equation.

Question

A particle is moving with velocity $\mathbf{v} = k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}})$, where $k$ is a constant. The general equation for its path is

Options:

(A) $y = x^2 + 	ext{constant}$

(B) $y^2 = x + 	ext{constant}$

(C) $xy = 	ext{constant}$

(D) $y^2 = x^2 + 	ext{constant}$

Sigma Percentile · Accepted Answer

### Unraveling the Path of a Particle: A Journey from Velocity to Trajectory

Imagine a particle moving through a two-dimensional plane. Its motion isn't simple or uniform; instead, its velocity depends entirely on its current coordinates. We are given a fascinating velocity vector field:

$$\mathbf{v} = k(y\hat{\mathbf{i}} + x\hat{\mathbf{j}})$$

Here, $k$ is just a constant. Our goal is to find the **general equation for its path**, also known as its trajectory. To do this, we need to strip away the element of time and find a direct relationship between the $y$ and $x$ coordinates.

### Breaking Down the Velocity

Any two-dimensional velocity vector can be expressed in terms of its horizontal and vertical components:

$$\mathbf{v} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}$$

By comparing this standard form with our given equation, we can extract the rates of change of the particle's position with respect to time $t$:

$$v_x = \frac{dx}{dt} = ky$$
$$v_y = \frac{dy}{dt} = kx$$

This tells us something interesting: the faster the particle moves in the $x$-direction, the higher it is on the $y$-axis, and vice versa.

### The Master Equation: Eliminating Time

To find the trajectory, we need an equation that only involves $y$ and $x$. We can achieve this by finding the slope of the path at any given point, which is $\frac{dy}{dx}$. By using the chain rule (or simply dividing the $y$-velocity by the $x$-velocity), we can eliminate the time differential $dt$:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{v_y}{v_x}$$

Substituting our components into this relation:

$$\frac{dy}{dx} = \frac{kx}{ky} = \frac{x}{y}$$

Notice how the constant $k$ elegantly cancels out! We are left with a beautiful, separable differential equation.

### Integrating to Find the Path

To solve this differential equation, we use the method of separation of variables. We group all the $y$ terms on one side and all the $x$ terms on the other:

$$y \, dy = x \, dx$$

Now, we integrate both sides:

$$\int y \, dy = \int x \, dx$$

Using the standard power rule for integration, we get:

$$\frac{y^2}{2} = \frac{x^2}{2} + C'$$

where $C'$ is our constant of integration.

### The Final Geometric Form

To clean up the equation, let's multiply everything by 2:

$$y^2 = x^2 + 2C'$$

Since $C'$ is an arbitrary constant, $2C'$ is also just an arbitrary constant. We can simply write it as `constant`:

$$y^2 = x^2 + 	ext{constant}$$

**Geometrically, what does this represent?** 
If we rearrange it as $y^2 - x^2 = 	ext{constant}$, we can recognize this as the standard equation for a family of **hyperbolas**. Depending on the initial starting position of the particle (which determines the value of the constant), the particle will trace out one of these hyperbolic curves. If the constant happens to be zero, the path degenerates into the straight lines $y = x$ or $y = -x$.

Animated Solution for Physics - Kinematics: A particle is moving with velocity $v = k (y \hat{i} + x \hat{j})$ , where $k$ is a constant. The general equation for its path is

Select Answer:

Visualized Solution

$v = v_{x} \hat{i} + v_{y} \hat{j}$

$v_{x} = \frac{d x}{d t}, v_{y} = \frac{d y}{d t}$

$\frac{d y}{d x} = \frac{v _{y}}{v _{x}}$

$y d y = x d x$

$\int y d y = \int x d x$

$y^{2} = x^{2} + constant$

The Sigma Insight: Motion in a Plane

Unraveling the Path of a Particle

Breaking Down the Velocity

The Master Equation

Integrating to Find the Path

The Final Geometric Form

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A particle is moving with velocity v=k(yi^+xj^​), where k is a constant. The general equation for its path is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=vx​i^+vy​j^​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vx​=dtdx​,vy​=dtdy​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​=vx​vy​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ydy=xdx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫ydy=∫xdx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=x2+constant