Question

If the radius of curvature (\(\rho\)) at (0, 1) of \(y = e^x\) is \(\alpha\sqrt{\beta}\), then \(\alpha^2+\beta\) is:

• A. 7
• B. 6
• C. 11
• D. 12

Hint:

The radius of curvature formula is a standard application of differentiation. Remember to evaluate the derivatives at the specified point *before* plugging them into the formula. The expression \(2^{3/2}\) can be tricky; remember that \(x^{a/b} = (\sqrt[b]{x})^a\), so \(2^{3/2} = (\sqrt{2})^3 = 2\sqrt{2}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The radius of curvature (\(\rho\)) of a curve \(y=f(x)\) at a given point measures the radius of a circle that best approximates the curve at that point. We need to calculate this value for \(y=e^x\) at the point (0, 1) and then use the given form to find \(\alpha\) and \(\beta\).
Step 2: Key Formula or Approach:
The formula for the radius of curvature is: \[ \rho = \frac{[1 + (y')^2]^{3/2}}{|y''|} \] where \(y'\) and \(y''\) are the first and second derivatives of \(y\) with respect to \(x\).
Step 3: Detailed Explanation:
1. Find the first and second derivatives of \(y = e^x\):
\[ y = e^x \] \[ y' = \frac{dy}{dx} = e^x \] \[ y'' = \frac{d^2y}{dx^2} = e^x \] 2. Evaluate the derivatives at the point (0, 1):
The point is given by x=0, y=1. \[ y'(0) = e^0 = 1 \] \[ y''(0) = e^0 = 1 \] 3. Calculate the radius of curvature \(\rho\) at (0, 1):
Substitute the values into the formula: \[ \rho = \frac{[1 + (y'(0))^2]^{3/2}}{|y''(0)|} = \frac{[1 + (1)^2]^{3/2}}{|1|} = \frac{[1+1]^{3/2}}{1} = 2^{3/2} \] \[ \rho = \sqrt{2^3} = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \] 4. Determine \(\alpha\) and \(\beta\):
We are given that the radius of curvature is in the form \(\alpha\sqrt{\beta}\).
Comparing \( \rho = 2\sqrt{2} \) with \( \alpha\sqrt{\beta} \), we have: \[ \alpha = 2, \quad \beta = 2 \] 5. Calculate \(\alpha^2 + \beta\):
\[ \alpha^2 + \beta = (2)^2 + 2 = 4 + 2 = 6 \] Step 4: Final Answer:
The value of \(\alpha^2 + \beta\) is 6.