Problem Analysis #42

[THOUGHT] We need to understand the problem step by step.

We have f(x) = (1/2)x^2 - x + ln(1+x), defined for x > -1 (since ln(1+x)). For s > 0, (s, f(s)) is a point on the curve.

We consider the perpendicular from (s, f(s)) to the y-axis. The foot of that perpendicular is the point on the y-axis with the same y-coordinate: (0, f(s)).

Now consider the tangent line to the curve at (s, f(s)). Let its equation be y = f'(s)(x - s) + f(s). The point where this tangent meets the y-axis is when x = 0: y = f'(s)(-s) + f(s) = f(s) - s f'(s).

So we have two points on the y-axis: A: (0, f(s)) (foot of perpendicular) B: (0, f(s) - s f'(s)) (tangent y-intercept)

The distance between these two points along the y-axis is the absolute difference in y-coordinates: |f(s) - (f(s) - s f'(s))| = |s f'(s)|.

Since s > 0, and f'(s) we can compute. But we need to see sign.

Given t is a positive real number. Let g(t) be the value of s such that the distance between those two points equals t. That is: |s f'(s)| = t.

But t > 0, so s f'(s) could be positive or negative? Let's compute f'(x).

f(x) = (1/2)x^2 - x + ln(1+x) f'(x) = x - 1 + 1/(1+x) = (x - 1) + 1/(1+x).

Simplify: f'(x) = (x - 1)(1+x)/(1+x) + 1/(1+x) = ((x-1)(x+1) + 1)/(1+x) = (x^2 - 1 + 1)/(1+x) = x^2/(1+x).

Indeed: x - 1 + 1/(1+x) = ( (x-1)(1+x) + 1 )/(1+x) = (x^2 - 1 + 1)/(1+x) = x^2/(1+x). So f'(x) = x^2/(1+x).

Thus s f'(s) = s * (s^2/(1+s)) = s^3/(1+s).

Since s > 0, s^3/(1+s) > 0. So distance = s^3/(1+s) = t.

Therefore, g(t) is the function that for given t > 0 returns s such that s^3/(1+s) = t.

We need to evaluate ∫_{1/2}^{27/4} g(t) dt.

Let’s denote s = g(t). Then t = s^3/(1+s). We can change variable: t = s^3/(1+s). Then dt = d/ds (s^3/(1+s)) ds.

Compute derivative: d/ds [s^3/(1+s)] = (3s^2(1+s) - s^3 * 1)/(1+s)^2 = (3s^2 + 3s^3 - s^3)/(1+s)^2 = (3s^2 + 2s^3)/(1+s)^2 = s^2(3 + 2s)/(1+s)^2.

Thus dt = s^2(3+2s)/(1+s)^2 ds.

When t = 1/2, we need corresponding s: solve s^3/(1+s) = 1/2 => 2s^3 = 1+s => 2s^3 - s - 1 = 0. Try s=1: 2-1-1=0. So s=1 is a root. Factor: (s-1)(2s^2+2s+1)=0. Quadratic discriminant negative. So s=1.

When t = 27/4 = 6.75, solve s^3/(1+s) = 27/4 => 4s^3 = 27(1+s) => 4s^3 - 27s - 27 = 0. Try s=3: 4*27 - 81 - 27 = 108 - 108 = 0. So s=3 is a root. Factor: (s-3)(4s^2+12s+9)=0? Let's check: (s-3)(4s^2+12s+9) = 4s^3+12s^2+9s -12s^2-36s-27 = 4s^3 -27s -27. Yes.

So s=3.

Thus integration bounds in s: from s=1 to s=3.

Now ∫{t=1/2}^{27/4} g(t) dt = ∫{s=1}^{3} s * (dt/ds) ds? Wait careful: ∫ g(t) dt = ∫ s dt. But dt = (s^2(3+2s)/(1+s)^2) ds. So ∫ s dt = ∫{s=1}^{3} s * (s^2(3+2s)/(1+s)^2) ds = ∫{1}^{3} s^3 (3+2s)/(1+s)^2 ds.

Thus we need to compute I = ∫_{1}^{3} s^3 (3+2s)/(1+s)^2 ds.

Simplify: Let u = s+1, then s = u-1. But maybe easier expand numerator.

Compute s^3(3+2s) = 3 s^3 + 2 s^4.

So integrand = (2 s^4 + 3 s^3)/(s+1)^2.

We can perform polynomial division or substitution.

Let's compute using Python symbolic integration.

We'll compute I exactly, then the problem asks: Evaluate ∫_{1/2}^{27/4} g(t) dt. Possibly result is something like p/q + ln 2. Then compute p+q.

Let's compute I numerically first to see approximate value.

[/THOUGHT] [PYTHON] import sympy as sp

s = sp.symbols('s') integrand = s**3 * (3 + 2*s) / (1 + s)**2 I = sp.integrate(integrand, (s, 1, 3)) print("Exact integral I:", I) print("Numerical value:", I.evalf()) [/PYTHON]

[THOUGHT] The exact integral is ( -\ln(2) + \ln(4) + \frac{157}{12} ). Simplify (\ln(4) = 2\ln(2)), so (-\ln(2) + 2\ln(2) = \ln(2)). Thus the result is (\frac{157}{12} + \ln(2)). This matches the form (\frac{p}{q} + \ln 2) with (p = 157), (q = 12). Then (p + q = 157 + 12 = 169).

[/THOUGHT] \boxed{169}