= Solution
Use the <Levi-Civita connection> and the <Riemann curvature tensor> convention
$$
\mathcal R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,
\qquad K_M(T,E)=\langle\mathcal R(E,T)T,E\rangle
$$
for orthonormal $T,E$. Thus the round <sphere> has positive <sectional curvature>. Work on a fixed <compact> parameter interval $[a,b]$ inside the given open interval; this keeps all differentiations and endpoint evaluations well-defined. The base <geodesic> is regular and has constant speed, so a constant rescaling of $t$ makes it unit-speed. Put $T=\partial_t\phi$, $J=\partial_s\phi$, $D_t=\nabla_T$, $D_s=\nabla_J$ and evaluate at $s=0$. The <torsion-free connection> gives $D_sT=D_tJ$ because the two parameter derivatives commute. The <geodesic> equation is $D_tT=0$.
Differentiating the <length of a curve> once and twice, using <metric compatibility>, gives
$$
L'(0)=\int_a^b\langle D_tJ,T\rangle\,dt
=[\langle J,T\rangle]_a^b,
$$
and
$$
L''(0)=\int_a^b\left(|D_tJ|^2-\langle D_tJ,T\rangle^2+\langle D_sD_tJ,T\rangle\right)dt.
$$
Let $A=D_sJ$. The curvature commutator gives $D_sD_tJ=D_tA+\mathcal R(J,T)J$, and the skew-adjoint curvature symmetry gives $\langle\mathcal R(J,T)J,T\rangle=-\langle\mathcal R(J,T)T,J\rangle$. Integration by parts therefore proves the <moving-endpoint second variation of Riemannian arc length>:
$$
\boxed{L''(0)=[\langle A,T\rangle]_a^b+
\int_a^b\left(|D_tJ|^2-\langle D_tJ,T\rangle^2-\langle\mathcal R(J,T)T,J\rangle\right)dt.}
$$
Equivalently, with $W=J-\langle J,T\rangle T$, the integral is $I(W,W)$, the <Riemannian index form>. The endpoints of this <geodesic variation> generally move, so the displayed boundary term must be retained. For a base speed $c$ instead of one, the boundary and integral have an overall factor $1/c$, and the subtracted square inside the integral is $\langle D_tJ,T\rangle^2/c^2$. Returning to the original parameter, every ruling has constant speed, so $L_{[a,b]}(s)=(b-a)|\partial_t\phi(0,s)|$ and the full open-interval length is $2\delta|\partial_t\phi(0,s)|$. Consequently
$$
\left.\frac{d^2}{ds^2}L(\gamma_s)\right|_{s=0}=\frac{2\delta}{b-a}L_{[a,b]}''(0).
$$
The preceding curvature formula, with the base-speed factor $1/c$ when necessary, therefore evaluates the original full length derivative using any compact subinterval. No extension to the missing endpoints is required.
Differentiating $D_tT=0$ in $s$ yields the <Jacobi field> equation
$$
D_t^2J+\mathcal R(J,T)T=0.
$$
The same ruling <curves> are <geodesics> of the surface with its <induced metric>, because the tangential part of their ambient acceleration is zero. Write its <second fundamental form> as $II$ and its induced <Levi-Civita connection> as $D^N$. In particular $II(T,T)=0$. The scalar $f=\langle J,T\rangle$ satisfies $f''=0$, by the <Jacobi field> equation, so $W=J-fT$ is a <Jacobi field> both intrinsically and in the ambient manifold. In a regular surface chart $h=|W|>0$, and $E=W/h$ is the perpendicular unit tangent. Along a two-dimensional <geodesic>, $D_t^NE=0$: its derivative is orthogonal to both $E$ and the parallel field $T$. The intrinsic <Jacobi field> equation consequently gives
$$
h''=-K_Nh.
$$
For the ambient derivative, the <Gauss formula> gives $D_tW=h'E+h\,II(T,E)$. Applying the same norm differentiation used above to $h=|W|$ gives
$$
hh''=|D_tW|^2-(h')^2-\langle\mathcal R(W,T)T,W\rangle
=h^2\bigl(\|II(T,E)\|^2-K_M(T,E)\bigr).
$$
Comparing these identities proves the <curvature comparison for a geodesically ruled surface>:
$$
\boxed{K_N=K_M(T,E)-\|II(T,E)\|^2\le K_M(T_pN).}
$$
This is also the <Gauss equation in a curved ambient manifold> with $II(T,T)=0$. It applies locally to each regular patch of the surface.
For strict inequality, take the Euclidean <embedded submanifold> parametrized by $\phi(t,s)=(t,s,ts)$. Each $t$-curve is a straight-line <geodesic>. Its <first fundamental form> has coefficients $E_0=1+s^2$, $F_0=st$, $G_0=1+t^2$; the normal $(-s,-t,1)/\sqrt{1+t^2+s^2}$ gives <second fundamental form> coefficients $e=0$, $f_0=1/\sqrt{1+t^2+s^2}$, $g_0=0$. Hence
$$
\boxed{K_N(t,s)=-\frac1{(1+t^2+s^2)^2}<0=K_{\mathbb R^3}.}
$$
Thus equality can fail everywhere, although every ruling is an ambient <geodesic>.
The length differentiation requires a nonconstant base ruling; a regular parametrized surface patch guarantees this. If “image of a smooth map” is read literally without a rank assumption, a degenerate member can fail even to have a first length derivative. For example, in the flat cylinder $\mathbb R\times S^1$, the map $\phi(t,s)=(st,e^{is})$ for $|t|<\delta$, $|s|<4\pi$ has <geodesic> rulings and a two-dimensional open image: every nonzero-$s$ point has full-rank parametrization, and the sole zero-$s$ image point also has a full-rank preimage at $s=2\pi,t=0$. Yet $L(\gamma_s)=2\delta|s|$ is not differentiable at zero. The variation formula therefore applies to nonconstant members, as usual. The curvature inequality still holds on the whole image: regular values are dense by <Sard theorem>, and both curvatures extend continuously to the remaining points.
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