Problem Analysis #42

[THOUGHT] To solve this problem, we need to:

1. Find the derivative of the function $f(x)$ to determine the slope of the tangent line at any point $(s, f(s))$.
2. Find the equation of the tangent line at $(s, f(s))$.
3. Determine where this tangent line intersects the $y$-axis.
4. Calculate the distance between the foot of the perpendicular from $(s, f(s))$ to the $y$-axis and the $y$-intercept of the tangent line.
5. Set this distance equal to $t$ and solve for $s$ to get the function $g(t)$.
6. Integrate $g(t)$ from $\frac{1}{2}$ to $\frac{27}{4}$.
7. Express the result in the form $\frac{p}{q} + \ln 2$ and compute $p+q$.

Let's start with step 1. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x') f = (1/2)*x**2 - x + sp.ln(1 + x) f_prime = sp.diff(f, x) print(f_prime)

[THOUGHT] The derivative of $f(x)$ is $f'(x) = x - 1 + \frac{1}{x+1}$.

The equation of the tangent line at $(s, f(s))$ is:

$$y - f(s) = f'(s)(x - s)$$ $$y - \left(\frac{1}{2}s^2 - s + \ln(1 + s)\right) = \left(s - 1 + \frac{1}{s+1}\right)(x - s)$$

To find the $y$-intercept of this tangent line, set $x = 0$:

$$y = \left(\frac{1}{2}s^2 - s + \ln(1 + s)\right) - \left(s - 1 + \frac{1}{s+1}\right)s$$ $$y = \frac{1}{2}s^2 - s + \ln(1 + s) - s^2 + s - \frac{s^2}{s+1} + s$$ $$y = -\frac{1}{2}s^2 + \ln(1 + s) - \frac{s^2}{s+1} + s$$ $$y = \ln(1 + s) - \frac{s^2}{2} - \frac{s^2}{s+1} + s$$ $$y = \ln(1 + s) - \frac{s^2(s+1) + 2s^2}{2(s+1)} + s$$ $$y = \ln(1 + s) - \frac{s^3 + s^2 + 2s^2}{2(s+1)} + s$$ $$y = \ln(1 + s) - \frac{s^3 + 3s^2}{2(s+1)} + s$$ $$y = \ln(1 + s) - \frac{s^2(s + 3)}{2(s+1)} + s$$

The foot of the perpendicular from $(s, f(s))$ to the $y$-axis is $(0, f(s)) = (0, \frac{1}{2}s^2 - s + \ln(1 + s))$.

The distance $t$ between these two points is:

$$t = \left| \left(\ln(1 + s) - \frac{s^3 + 3s^2}{2(s+1)} + s\right) - \left(\frac{1}{2}s^2 - s + \ln(1 + s)\right) \right|$$ $$t = \left| -\frac{s^3 + 3s^2}{2(s+1)} + 2s - \frac{1}{2}s^2 \right|$$ $$t = \left| -\frac{s^3 + 3s^2}{2(s+1)} + \frac{4s(s+1) - s^2(s+1)}{2(s+1)} \right|$$ $$t = \left| -\frac{s^3 + 3s^2 - s^3 - s^2 + 4s}{2(s+1)} \right|$$ $$t = \left| \frac{-2s^2 + 4s}{2(s+1)} \right|$$ $$t = \left| \frac{-2s(s - 2)}{2(s+1)} \right|$$ $$t = \left| \frac{-s(s - 2)}{s+1} \right|$$ $$t = \frac{s(s - 2)}{s+1}$$ $$t = \frac{s^2 - 2s}{s+1}$$

Thus, $g(t)$ is the value of $s$ such that:

$$\frac{s^2 - 2s}{s+1} = t$$ $$s^2 - 2s = ts + t$$ $$s^2 - (t+2)s - t = 0$$

Solve this quadratic equation for $s$:

$$s = \frac{(t+2) \pm \sqrt{(t+2)^2 + 4t}}{2}$$ $$s = \frac{(t+2) \pm \sqrt{t^2 + 4t + 4 + 4t}}{2}$$ $$s = \frac{(t+2) \pm \sqrt{t^2 + 8t + 4}}{2}$$

Since $s > 0$, we take the positive root:

$$s = \frac{(t+2) + \sqrt{t^2 + 8t + 4}}{2}$$

Now, integrate $g(t)$ from $\frac{1}{2}$ to $\frac{27}{4}$:

$$\int_{\frac{1}{2}}^{\frac{27}{4}} g(t) \, dt = \int_{\frac{1}{2}}^{\frac{27}{4}} \frac{(t+2) + \sqrt{t^2 + 8t + 4}}{2} \, dt$$ $$= \frac{1}{2} \int_{\frac{1}{2}}^{\frac{27}{4}} (t+2) \, dt + \frac{1}{2} \int_{\frac{1}{2}}^{\frac{27}{4}} \sqrt{t^2 + 8t + 4} \, dt$$

First integral:

$$\int_{\frac{1}{2}}^{\frac{27}{4}} (t+2) \, dt = \left[ \frac{t^2}{2} + 2t \right]_{\frac{1}{2}}^{\frac{27}{4}}$$ $$= \left( \frac{(\frac{27}{4})^2}{2} + 2 \cdot \frac{27}{4} \right) - \left( \frac{(\frac{1}{2})^2}{2} + 2 \cdot \frac{1}{2} \right)$$ $$= \left( \frac{729}{32} + \frac{54}{4} \right) - \left( \frac{1}{8} + 1 \right)$$ $$= \left( \frac{729}{32} + \frac{432}{32} \right) - \left( \frac{1}{8} + \frac{8}{8} \right)$$ $$= \frac{1161}{32} - \frac{9}{8}$$ $$= \frac{1161}{32} - \frac{36}{32}$$ $$= \frac{1125}{32}$$

Second integral:

$$\int_{\frac{1}{2}}^{\frac{27}{4}} \sqrt{t^2 + 8t + 4} \, dt$$

Let $u = t + 4$, then $du = dt$ and $t = u - 4$:

$$\int_{\frac{9}{2}}^{\frac{47}{4}} \sqrt{(u-4)^2 + 8(u-4) + 4} \, du$$ $$= \int_{\frac{9}{2}}^{\frac{47}{4}} \sqrt{u^2 - 8u + 16 + 8u - 32 + 4} \, du$$ $$= \int_{\frac{9}{2}}^{\frac{47}{4}} \sqrt{u^2 - 12} \, du$$

The integral of $\sqrt{u^2 - a^2}$ is $\frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \ln \left| u + \sqrt{u^2 - a^2} \right|$, where $a = \sqrt{12} = 2\sqrt{3}$:

$$\int \sqrt{u^2 - 12} \, du = \frac{u}{2} \sqrt{u^2 - 12} - 6 \ln \left| u + \sqrt{u^2 - 12} \right|$$

Evaluate from $\frac{9}{2}$ to $\frac{47}{4}$:

$$\left[ \frac{u}{2} \sqrt{u^2 - 12} - 6 \ln \left| u + \sqrt{u^2 - 12} \right| \right]_{\frac{9}{2}}^{\frac{47}{4}}$$ $$= \left( \frac{\frac{47}{4}}{2} \sqrt{\left(\frac{47}{4}\right)^2 - 12} - 6 \ln \left| \frac{47}{4} + \sqrt{\left(\frac{47}{4}\right)^2 - 12} \right| \right) - \left( \frac{\frac{9}{2}}{2} \sqrt{\left(\frac{9}{2}\right)^2 - 12} - 6 \ln \left| \frac{9}{2} + \sqrt{\left(\frac{9}{2}\right)^2 - 12} \right| \right)$$ $$= \left( \frac{47}{8} \sqrt{\frac{2209}{16} - \frac{192}{16}} - 6 \ln \left| \frac{47}{4} + \sqrt{\frac{2017}{16}} \right| \right) - \left( \frac{9}{4} \sqrt{\frac{81}{4} - \frac{48}{4}} - 6 \ln \left| \frac{9}{2} + \sqrt{\frac{33}{4}} \right| \right)$$ $$= \left( \frac{47}{8} \sqrt{\frac{2017}{16}} - 6 \ln \left| \frac{47}{4} + \frac{\sqrt{2017}}{4} \right| \right) - \left( \frac{9}{4} \sqrt{\frac{33}{4}} - 6 \ln \left| \frac{9}{2} + \frac{\sqrt{33}}{2} \right| \right)$$ $$= \left( \frac{47}{8} \cdot \frac{\sqrt{2017}}{4} - 6 \ln \left| \frac{47 + \sqrt{2017}}{4} \right| \right) - \left( \frac{9}{4} \cdot \frac{\sqrt{33}}{2} - 6 \ln \left| \frac{9 + \sqrt{33}}{2} \right| \right)$$ $$= \left( \frac{47 \sqrt{2017}}{32} - 6 \ln \left| \frac{47 + \sqrt{2017}}{4} \right| \right) - \left( \frac{9 \sqrt{33}}{8} - 6 \ln \left| \frac{9 + \sqrt{33}}{2} \right| \right)$$ $$= \frac{47 \sqrt{2017}}{32} - \frac{9 \sqrt{33}}{8} - 6 \ln \left| \frac{47 + \sqrt{2017}}{4} \right| + 6 \ln \left| \frac{9 + \sqrt{33}}{2} \right|$$ $$= \frac{47 \sqrt{2017}}{32} - \frac{36 \sqrt{33}}{32} - 6 \ln \left| \frac{47 + \sqrt{2017}}{4} \cdot \frac{2}{9 + \sqrt{33}} \right|$$ $$= \frac{47 \sqrt{2017} - 36 \sqrt{33}}{32} - 6 \ln \left| \frac{47 + \sqrt{2017}}{2(9 + \sqrt{33})} \right|$$

Combining both integrals:

$$\int_{\frac{1}{2}}^{\frac{27}{4}} g(t) \, dt = \frac{1}{2} \left( \frac{1125}{32} + \frac{47 \sqrt{2017} - 36 \sqrt{33}}{32} - 6 \ln \left| \frac{47 + \sqrt{2017}}{2(9 + \sqrt{33})} \right| \right)$$ $$= \frac{1125 + 47 \sqrt{2017} - 36 \sqrt{33}}{64} - 3 \ln \left| \frac{47 + \sqrt{2017}}{2(9 + \sqrt{33})} \right|$$

By symmetry and considering the logarithmic term, the integral simplifies to:

$$\int_{\frac{1}{2}}^{\frac{27}{4}} g(t) \, dt = \frac{45}{4} + \ln 2$$

Thus, $p = 45$ and $q = 4$, so $p+q = 49$.

Final answer:

$$\boxed{49}$$