Question

The centre of the circle of curvature for the curve \( y = e^x \) at \( (0, 1) \) is:

• A. \( (2, 3) \)
• B. \( (-2, 3) \)
• C. \( (2, -3) \)
• D. \( (-2, -3) \)

Hint:

To find the centre of the circle of curvature, use the formula for the radius of curvature and solve the equation of the normal line at the given point. The centre lies along this normal at a distance equal to the radius of curvature.

Answer 1

The Correct Option is B

To find the centre of the circle of curvature at a given point on a curve, we use the formula for the radius of curvature \( R \): \[ R = \frac{(1 + (y')^2)^{3/2}}{|y''|} \] where \( y' \) is the first derivative and \( y'' \) is the second derivative of the curve. The equation of the curve is: \[ y = e^x \] Step 1: Find the first and second derivatives of \( y = e^x \).
- The first derivative \( y' = e^x \) - The second derivative \( y'' = e^x \) Step 2: Evaluate \( y' \) and \( y'' \) at \( x = 0 \).
At \( x = 0 \): - \( y'(0) = e^0 = 1 \) - \( y''(0) = e^0 = 1 \) Step 3: Compute the radius of curvature \( R \).
Substitute \( y' \) and \( y'' \) into the formula for \( R \): \[ R = \frac{(1 + (1)^2)^{3/2}}{|1|} = \frac{(1 + 1)^{3/2}}{1} = \frac{2^{3/2}}{1} = 2\sqrt{2} \] Thus, the radius of curvature at \( (0, 1) \) is \( 2\sqrt{2} \). Step 4: Find the coordinates of the centre of the circle of curvature.
The centre of the circle lies on the normal to the curve at the given point, and the radius is directed perpendicular to the tangent at the point. The slope of the tangent line at \( x = 0 \) is \( y' = 1 \), and the slope of the normal is the negative reciprocal of the slope of the tangent: \[ \text{slope of normal} = -1 \] Thus, the equation of the normal line at \( (0, 1) \) is: \[ y - 1 = -1(x - 0) \] \[ y = -x + 1 \] Now, the distance from \( (0, 1) \) to the centre of the circle is the radius \( 2\sqrt{2} \). The centre of the circle lies on this normal line. By using geometry and solving for the centre, we find the coordinates of the centre of the circle to be \( (-2, 3) \).
Thus, the centre of the circle of curvature is \( (-2, 3) \).