Finding Radius of Curvature for $y = e^x$

The question asks us to find the radius of curvature for the function $y = e^x$ at the specific point $(0, 1)$. We are told this radius of curvature, let's call it $P$, is equal to $\alpha \sqrt{\beta}$. Our goal is to calculate the value of the expression $\alpha^2 + \beta$.

Understanding Radius of Curvature

The radius of curvature ($P$) at a point on a curve measures how tightly the curve is bending at that point. For a function $y = f(x)$, the formula to calculate the radius of curvature is:

$ P = \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|} $

To use this formula, we need to find the first and second derivatives of the given function and evaluate them at the specified point.

Step 1: Calculate Derivatives

First, let's find the first and second derivatives of the function $y = e^x$.

• First Derivative ($\frac{dy}{dx}$): Differentiating $y = e^x$ with respect to $x$, we get: $ \frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x $
• Second Derivative ($\frac{d^2y}{dx^2}$): Differentiating the first derivative ($\frac{dy}{dx} = e^x$) with respect to $x$, we get: $ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) = e^x $

Step 2: Evaluate Derivatives at Point $(0, 1)$

We need to evaluate these derivatives at the point $(0, 1)$. This means substituting $x=0$ into our derivative expressions.

• Value of $\frac{dy}{dx}$ at $x=0$: $ \frac{dy}{dx}\bigg|_{x=0} = e^0 = 1 $
• Value of $\frac{d^2y}{dx^2}$ at $x=0$: $ \frac{d^2y}{dx^2}\bigg|_{x=0} = e^0 = 1 $

Step 3: Calculate the Radius of Curvature ($P$)

Now we plug the evaluated derivatives into the radius of curvature formula:

$ P = \frac{\left(1 + \left(\frac{dy}{dx}\bigg|_{x=0}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\bigg|_{x=0}\right|} $

Substitute the values we found:

$ P = \frac{\left(1 + (1)^2\right)^{3/2}}{|1|} $

$ P = \frac{(1 + 1)^{3/2}}{1} $

$ P = (2)^{3/2} $

To simplify $(2)^{3/2}$, we can write it as $2^{1 + 1/2} = 2^1 \cdot 2^{1/2} = 2\sqrt{2}$.

So, the radius of curvature is:

$ P = 2\sqrt{2} $

Step 4: Find $\alpha$, $\beta$, and Calculate $\alpha^2 + \beta$

The problem states that the radius of curvature is $\alpha \sqrt{\beta}$. We found that $P = 2\sqrt{2}$.

By comparing $2\sqrt{2}$ with $\alpha \sqrt{\beta}$, we can determine the values of $\alpha$ and $\beta$:

• $\alpha = 2$
• $\beta = 2$

Finally, we calculate $\alpha^2 + \beta$:

$ \alpha^2 + \beta = (2)^2 + 2 $

$ \alpha^2 + \beta = 4 + 2 $

$ \alpha^2 + \beta = 6 $

Final Result

The calculated value for $\alpha^2 + \beta$ is 6.