91Ó°ÊÓ

In physics, the displacement function describes the change in position of an object over time. It is a vital component in understanding motion. In our original exercise, the displacement as a function of time is given by the equation \(x^3 = t^3 + 1\). This equation connects the position \(x\) with time \(t\).
- To find the position \(x\) at any given moment, one must solve this equation for \(x\) in terms of \(t\). - The displacement function encapsulates how far a point or object has moved from its starting position at a given time.
Understanding it helps us derive other characteristics like velocity and acceleration, which are subsequent applications of these foundational relations.

Velocity represents the rate of change of displacement with respect to time. It tells us how fast and in what direction an object is moving. To determine the velocity from the displacement function, we need to differentiate the displacement with respect to time.
- Given the displacement equation \(x^3 = t^3 + 1\), the differentiation requires special techniques such as implicit differentiation.- Differentiating both sides with respect to time gives: \[ \frac{d}{dt}(x^3) = \frac{d}{dt}(t^3 + 1) \] Using the chain rule of calculus here helps in finding the expression for velocity. We find: \[ 3x^2 \frac{dx}{dt} = 3t^2 \] By solving this equation, we arrive at the velocity function: \[ v = \frac{dx}{dt} = \frac{t^2}{x^2} \]
This velocity calculation demonstrates how differentiation helps us understand changes in motion direction and speed at any given point in time.

Implicit differentiation is a powerful technique when dealing with functions that are not conveniently solved or expressed in terms of a single variable. It is often used when variables are intertwined, as in our displacement function \(x^3 = t^3 + 1\).
- Here, differentiating implicitly allows us not to solve \(x\) directly for \(t\), which in some cases can be complicated or impossible.- By differentiating both sides of \(x^3 = t^3 + 1\) with respect to \(t\), and applying chain rule methods, we handle the differentiation smoothly: \[ 3x^2 \frac{dx}{dt} = 3t^2 \]
Implicit differentiation helps uncover the relationships between the rate at which \(x\) and \(t\) change, allowing us to derive important quantities such as velocity and later acceleration.

The quotient rule is a critical calculus technique used to differentiate functions that are the ratio of two other functions. This method was necessary again in the final step of finding the acceleration by differentiating the velocity function, which is given by \(v = \frac{t^2}{x^2}\).
- To find the acceleration (the second derivative of the displacement), we apply the quotient rule: \[ \frac{d}{dt} \left( \frac{t^2}{x^2} \right) = \frac{x^2 \frac{d}{dt}(t^2) - t^2 \frac{d}{dt}(x^2)}{(x^2)^2} \]- In this calculation, each part is carefully handled: - \(\frac{d}{dt}(t^2) = 2t\) is straightforward. - For \(\frac{d}{dt}(x^2)\), we make use of the chain rule, since \(x\) is a function of \(t\).
The result of the differentiation provides: \[ = \frac{2tx^2 - 2t^4/x^2}{x^4} = \frac{2t}{x^4} \]This reveals the acceleration, utilizing skills from both implicit differentiation and the quotient rule efficiently.