91Ó°ÊÓ

In differential calculus, the right-hand derivative focuses on examining the rate of change of a function as the input approaches a given point from the positive side. For a function, say, \( g(t) = f(\sqrt{t}) \) defined for \( t \geq 0 \), the right-hand derivative at \( t = 0 \) is given by:
\[ g_{+}^{\prime}(0) = \lim_{h \rightarrow 0^{+}} \frac{g(h) - g(0)}{h} \]
Here, we're interested in how \(g\) behaves as \(t\) approaches 0 from larger values. By employing Taylor’s Theorem, we approximate \(f(x)\) near 0. Suppose \( f: \mathbb{R} \rightarrow \mathbb{R} \) satisfies \( f^{\prime}(0) = 0 \). Then, Taylor’s expansion becomes:
\[ f(x) = f(0) + \frac{f''(0)}{2}x^2 + o(x^2) \]
Substituting \( x = \sqrt{t} \) into \(f(x)\) to get \( g(t) = f(\sqrt{t}) \), we derive:
\[ g(t) = f(0) + \frac{f''(0)}{2}t + o(t) \]
To find \( g_{+}^{\prime}(0) \):
\[ g_{+}^{\prime}(0) = \lim_{h \rightarrow 0^{+}} \frac{g(h) - g(0)}{h} = \lim_{h \rightarrow 0^{+}} \frac{ f(0) + \frac{f''(0)}{2}h + o(h) - f(0)}{h} = \frac{f''(0)}{2} \]
This shows the right-hand derivative of \( g \) at 0.

Taylor’s Theorem is a crucial tool in differential calculus that allows us to approximate functions near a given point using polynomials. It's particularly useful in providing insights into the behavior of functions through their derivatives. Given a function \( f \) that is sufficiently smooth, Taylor's Theorem states:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) \]
Here, \( R_n(x) \) represents the remainder term. In practice, we often truncate the series to a manageable number of terms, relying on higher-order terms only when higher precision is required.
In our exercise, we simplified the Taylor expansion around 0 since \( f^{\text{prime}}(0) = 0 \):
\[ f(x) \approx f(0) + \frac{f''(0)}{2}x^2 \]
This simple form provided a straightforward path to deducing the behavior of the transformed function \( g(t) = f(\sqrt{t}) \). Substituting and differentiating led us to the result we sought for the right-hand derivative.

In geometry and calculus, a tangent vector represents the direction and rate of change of a curve at a particular point. Understanding the tangent vector is crucial for describing the dynamic behavior of curves. Suppose \( c: \mathbb{R} \rightarrow M \) is a smooth curve satisfying \( c^{\prime}(0) = 0 \in M \), and we define \( \gamma(t) = c(\sqrt{t}) \) for \( t \geq 0 \). We aim to link \( c''(0) \) and \( \gamma'(0) \).
First, noting that \( \gamma(t) = c(\sqrt{t}) \), we use the chain rule to find \( \gamma'(0) \):
\[ \gamma'(0) = \lim_{h \rightarrow 0^{+}} \frac{\gamma(h) - \gamma(0)}{h} = \lim_{h \rightarrow 0^{+}} \frac{c(\sqrt{h}) - c(0)}{h} \]
Applying the Taylor expansion for \( c \) around 0, knowing \(c'(0)=0\) simplifies to:
\[ c(x) = c(0) + \frac{c''(0)}{2}x^2 + o(x^2) \]
Substituting \( x = \sqrt{h} \) results in:
\[ \gamma(h) = c(\sqrt{h}) = c(0) + \frac{c''(0)}{2}h + o(h) \]
Thus:
\[ \gamma'(0) = \lim_{h \rightarrow 0^{+}} \frac{c(0) + \frac{c''(0)}{2}h + o(h) - c(0)}{h} = \frac{c''(0)}{2} \]
Hence, we've shown \( c''(0) = 2 \gamma'(0) \). Understanding tangent vectors this way helps us grasp the geometric properties of curves in multivariate calculus.