A particle is moving along the parabola y²=12x at the uniform rate of 10cm/s.Find the components of velocity parallel to each of the axes when particle is at point (3,6).

To find the components of velocity parallel to the x-axis and y-axis for a particle moving along the parabola defined by the equation \(y^2 = 12x\) at the point (3, 6), we need to first understand the relationship between the position of the particle and its velocity. The particle is moving at a uniform speed of 10 cm/s, and we can use calculus to determine the velocity components.

Understanding the Parabola

The equation \(y^2 = 12x\) describes a parabola that opens to the right. To analyze the motion of the particle, we can express \(y\) in terms of \(x\) as follows:

y = ±√(12x)

At the point (3, 6), we can confirm that \(6^2 = 12 \cdot 3\), which holds true. Thus, the particle is indeed on the parabola.

Finding the Derivative

To determine the velocity components, we first need to find the derivative of \(y\) with respect to \(x\). This will help us understand how \(y\) changes as \(x\) changes:

Using implicit differentiation on \(y^2 = 12x\), we differentiate both sides:

• 2y(dy/dx) = 12

From this, we can solve for \(dy/dx\):

dy/dx = 12/(2y) = 6/y

At the point (3, 6), substituting \(y = 6\) gives:

dy/dx = 6/6 = 1

Velocity Components

Now that we have the slope of the tangent to the parabola at the point (3, 6), we can find the components of the velocity. The total velocity \(v\) of the particle is given as 10 cm/s. The components of the velocity can be expressed as:

• v_x = velocity component along the x-axis
• v_y = velocity component along the y-axis

Using the relationship between the components of velocity and the slope:

v_y = (dy/dx) * v_x

Let \(v_x\) be the component along the x-axis. Since the particle is moving at a uniform speed, we can express the total velocity as:

v = √(v_x² + v_y²)

Substituting \(v_y\) into this equation gives:

10 = √(v_x² + (dy/dx * v_x)²)

Substituting \(dy/dx = 1\):

10 = √(v_x² + v_x²) = √(2v_x²)

Squaring both sides results in:

100 = 2v_x²

Solving for \(v_x\):

v_x² = 50

v_x = √50 = 5√2 cm/s

Calculating v_y

Now we can find \(v_y\) using the relationship we established earlier:

v_y = (dy/dx) * v_x = 1 * 5√2 = 5√2 cm/s

Final Results

Thus, the components of the velocity of the particle at the point (3, 6) are:

• v_x = 5√2 cm/s
• v_y = 5√2 cm/s

These values indicate how fast the particle is moving in both the x and y directions as it travels along the parabola. This method illustrates the application of calculus in understanding motion along a curve.