To find the $y$-coordinate of the center of curvature for the curve given by the parametric equations:

$x = ct \quad \text{and} \quad y = \frac{c}{t}$

we need to use the formula for the coordinates of the center of curvature $(x_c, y_c)$ in parametric form. The center of curvature is located at a point where the normal to the curve at a given point intersects the osculating circle. The formula for the center of curvature is given by:

$y_c = y + \frac{\left( \dot{x}^2 + \dot{y}^2 \right)^{3/2}}{\dot{x} \ddot{y} - \dot{y} \ddot{x}}$

Where:

• $\dot{x} = \frac{dx}{dt}$
• $\dot{y} = \frac{dy}{dt}$
• $\ddot{x} = \frac{d^2x}{dt^2}$
• $\ddot{y} = \frac{d^2y}{dt^2}$

Step 1: Compute the first and second derivatives of $x(t)$ and $y(t)$

Given $x = ct$ and $y = \frac{c}{t}$, let's find the derivatives:

$\dot{x} = \frac{dx}{dt} = c$ $\dot{y} = \frac{dy}{dt} = -\frac{c}{t^2}$

Now the second derivatives:

$\ddot{x} = \frac{d^2x}{dt^2} = 0$ $\ddot{y} = \frac{d^2y}{dt^2} = \frac{2c}{t^3}$

Step 2: Apply the formula for $y_c$

Now substitute these values into the formula for $y_c$:

$y_c = \frac{c}{t} + \frac{\left( c^2 + \left(-\frac{c}{t^2}\right)^2 \right)^{3/2}}{c \cdot \frac{2c}{t^3} - \left( -\frac{c}{t^2} \right) \cdot 0}$

Simplifying:

$y_c = \frac{c}{t} + \frac{\left( c^2 + \frac{c^2}{t^4} \right)^{3/2}}{2c^2/t^3}$

Simplify further:

$y_c = \frac{c}{t} + \frac{c^3 \left( 1 + \frac{1}{t^4} \right)^{3/2}}{2c^2/t^3}$

$y_c = \frac{c}{t} + \frac{t^3 \cdot c \left( 1 + \frac{1}{t^4} \right)^{3/2}}{2}$

This is the expression for the $y$-coordinate of the center of curvature.

Would you like more details or have any questions?

5 Related Questions:

1. How do we interpret the center of curvature geometrically?
2. What is the significance of curvature in parametric equations?
3. Can this method be applied to non-parametric equations as well?
4. How does the center of curvature relate to the radius of curvature?
5. How do we find the radius of curvature for a given parametric curve?

Tip:

For curves given in parametric form, always compute derivatives carefully as they are key to finding curvature and related properties like the center of curvature.