Problem 8 - The Calculus with Analytic Geometry

Question

A particle is moving along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). Find the instantaneous velocity \(v\left(t_{1}\right) \mathrm{ft} / \mathrm{sec}\) at \(t_{1} \mathrm{sec}\); and then find \(v\left(t_{1}\right)\) for the particular value of \(t_{1}\) given. $$ s=\sqrt[3]{t+2} ; t_{1}=6 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \( t_1 = 6 \) seconds is \( \frac{1}{12} \, ft/s \).

Step 1: Understand the Problem

We need to find the instantaneous velocity of a particle at a given time. The position of the particle is given by the equation of motion, \[ s = \root{3}{t + 2} \], and we want to find the velocity at time \( t_1 = 6 \) seconds.

Step 2: Find the Velocity Function

The instantaneous velocity is the derivative of the position function with respect to time. Let's take the derivative of \[ s = \root{3}{t + 2} \] with respect to time \( t \).

Step 3: Apply the Chain Rule

Rewrite the position function in a more convenient form for differentiation: \[ s = (t + 2)^{\frac{1}{3}} \]. Use the chain rule to differentiate this:\[ \frac{ds}{dt} = \frac{d}{dt} (t + 2)^{\frac{1}{3}} \]. This gives us:\[ \frac{ds}{dt} = \frac{1}{3} (t + 2)^{-\frac{2}{3}} \times \frac{d}{dt}(t+2) \]. Since \[ \frac{d}{dt}(t + 2) = 1 \], the velocity function is:\[ v(t) = \frac{1}{3}(t + 2)^{-\frac{2}{3}} \].

Step 4: Substitute the Given Time

We need to find the velocity at \( t_1 = 6 \). Substitute \( t_1 = 6 \) into the velocity function:\[ v(6) = \frac{1}{3}(6 + 2)^{-\frac{2}{3}} \]. Thus:\[ v(6) = \frac{1}{3}(8)^{-\frac{2}{3}} \].

Step 5: Simplify

Simplify the expression from the previous step:\[ (8)^{-\frac{2}{3}} = \frac{1}{(8)^{\frac{2}{3}}} = \frac{1}{(\root{3}{8})^2} = \frac{1}{2^2} = \frac{1}{4} \]. This gives us:\[ v(6) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \, ft/s \]

Key Concepts

differentiation using chain rule

Differentiation is a key concept in calculus. It helps us understand how a function changes at any point. When we need to differentiate a composite function, the chain rule is our tool of choice. In this exercise, we start with the position function of the particle: $$ s = \root{3}{t + 2} $$. To apply the chain rule, we first rewrite it using exponents: $$ s = (t + 2)^{\frac{1}{3}} $$. The chain rule states that if we have a function such as $$ s = g(f(t)) $$ then the derivative of s with respect to t is: $$ \frac{ds}{dt} = g'(f(t)) \times f'(t) $$. In our problem, $$ g(u) = u^{\frac{1}{3}} $$ and $$ f(t) = t + 2 $$. So, we find the derivatives of these inner and outer functions. First: $$ \frac{d}{dt} (t + 2) = 1 $$. And then: $$ \frac{d}{du} (u^{\frac{1}{3}}) = \frac{1}{3} u^{-\frac{2}{3}} $$. By substituting back, we get: $$ \frac{ds}{dt} = \frac{1}{3} (t + 2)^{-\frac{2}{3}} \times 1 = \frac{1}{3} (t + 2)^{-\frac{2}{3}} $$.

equation of motion

Equations of motion describe the behavior of an object over time. They help us predict future positions and calculate velocities. In our exercise, we're given the position of a particle as: $$ s = \root{3}{t + 2} $$. This equation tells us how far the particle is from point O at any time t. To find instantaneous velocity, we need the rate of change of this position function. The derivative of the position function with respect to time gives us this rate of change, known as velocity. Instantaneous velocity captures the object's speed and direction at a specific moment. It essentially tells us how fast and in what direction the particle is moving at any point in time t.

simplification of derivatives

Simplifying derivatives is crucial for obtaining meaningful results. After differentiating our position function $$ s = (t + 2)^{\frac{1}{3}} $$ to get: $$ \frac{ds}{dt} = \frac{1}{3} (t + 2)^{-\frac{2}{3}} $$, we must simplify it for practical use. Let's substitute $$ t = 6 $$ into our derivative to find the instantaneous velocity: $$ v(6) = \frac{1}{3} (8)^{-\frac{2}{3}} $$.Breaking it down: since $$ 8 = 2^3$$, we have: $$ (8)^{-\frac{2}{3}} = \frac{1}{(8)^{\frac{2}{3}}} = \frac{1}{(2^3)^{\frac{2}{3}}} = \frac{1}{2^2} = \frac{1}{4}$$.Thus, $$ v(6) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \text{ ft/s} $$.Simplifying derivatives often involves breaking down exponents and using basic arithmetic. This ensures our final velocity value is clear and accurate.