Key Concepts and Formulas

• Geometric Interpretation of Derivative: The derivative of a function $f(x)$ at a point $x_0$, denoted by $f'(x_0)$, represents the slope of the tangent line to the curve $y=f(x)$ at the point $(x_0, f(x_0))$. If the tangent line makes an angle $\theta$ with the positive x-axis, then $f'(x_0) = \tan \theta$.
• Substitution Rule for Definite Integrals: To evaluate an integral of the form $\int_a^b g(h(t)) h'(t) dt$, we can use the substitution $z = h(t)$. Then $dz = h'(t) dt$. The limits of integration are changed accordingly: the lower limit becomes $h(a)$ and the upper limit becomes $h(b)$. The integral transforms to $\int_{h(a)}^{h(b)} g(z) dz$.
• Power Rule for Integration: For any real number $n \neq -1$, $\int z^n dz = \frac{z^{n+1}}{n+1} + C$.

Step-by-Step Solution

Step 1: Determine the values of the derivative at the given points. The problem states that the tangent to the curve $y=f(x)$ at $(1, f(1))$ makes an angle of $\pi/6$ with the positive x-axis. Using the geometric interpretation of the derivative, the slope of the tangent at $x=1$ is $f'(1)$. Therefore, $f'(1) = \tan(\pi/6) = \frac{1}{\sqrt{3}}$. Similarly, the tangent at $(3, f(3))$ makes an angle of $\pi/4$ with the positive x-axis. Thus, $f'(3) = \tan(\pi/4) = 1$.

Step 2: Set up the integral and apply the substitution. We are asked to evaluate 27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t. Let's use the substitution $z = f'(t)$. Differentiating with respect to $t$, we get $dz = f''(t) dt$. Now, we need to change the limits of integration: When $t=1$, $z = f'(1) = \frac{1}{\sqrt{3}}$. When $t=3$, $z = f'(3) = 1$. The integral becomes: 27 \int_\limits{1/\sqrt{3}}^1 \left(z^2+1\right) dz

Step 3: Evaluate the transformed definite integral. Using the power rule for integration, we integrate $(z^2+1)$ with respect to $z$: $\int \left(z^2+1\right) dz = \frac{z^3}{3} + z + C$ Now, we apply the limits of integration: 27 \left[\frac{z^3}{3} + z\right]_\limits{1/\sqrt{3}}^1 = 27 \left[\left(\frac{1^3}{3} + 1\right) - \left(\frac{(1/\sqrt{3})^3}{3} + \frac{1}{\sqrt{3}}\right)\right] $= 27 \left[\left(\frac{1}{3} + 1\right) - \left(\frac{1/(3\sqrt{3})}{3} + \frac{1}{\sqrt{3}}\right)\right]$ $= 27 \left[\frac{4}{3} - \left(\frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}}\right)\right]$ $= 27 \left[\frac{4}{3} - \left(\frac{1}{9\sqrt{3}} + \frac{9}{9\sqrt{3}}\right)\right]$ $= 27 \left[\frac{4}{3} - \frac{10}{9\sqrt{3}}\right]$

Step 4: Simplify the expression to the form $\alpha + \beta \sqrt{3}$. $27 \left[\frac{4}{3} - \frac{10}{9\sqrt{3}}\right] = 27 \times \frac{4}{3} - 27 \times \frac{10}{9\sqrt{3}}$ $= 9 \times 4 - 3 \times \frac{10}{\sqrt{3}}$ $= 36 - \frac{30}{\sqrt{3}}$ To rationalize the term with $\sqrt{3}$: $= 36 - \frac{30\sqrt{3}}{3}$ $= 36 - 10\sqrt{3}$ The expression is in the form $\alpha + \beta \sqrt{3}$, where $\alpha = 36$ and $\beta = -10$. Both $\alpha$ and $\beta$ are integers.

Step 5: Calculate $\alpha + \beta$. We need to find the value of $\alpha + \beta$. $\alpha + \beta = 36 + (-10) = 36 - 10 = 26$

Common Mistakes & Tips

• Incorrectly changing limits of integration: Always ensure that the new limits for the substituted variable correspond to the original limits of the integration variable.
• Algebraic errors in simplification: Be meticulous with fractions and radicals, especially when rationalizing denominators.
• Forgetting the constant factor: The constant 27 must be applied to the entire result of the definite integral.

Summary

The problem was solved by first using the geometric interpretation of the derivative to find the values of $f'(1)$ and $f'(3)$. Then, a substitution was applied to the definite integral, transforming it into a simpler polynomial integral. The transformed integral was evaluated using the power rule for integration and the new limits. Finally, the result was simplified and expressed in the form $\alpha + \beta \sqrt{3}$ to identify $\alpha$ and $\beta$, and their sum was calculated.

The final answer is $\boxed{26}$.