Problem 7 - 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. $$ s=\frac{2}{\sqrt{5 t+6}} ; t_{1}=2 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \( t = 2 \) is \( v(2) = -\frac{5}{64} \text{ ft/sec} \).

Step 1 - Find the expression for instantaneous velocity

The instantaneous velocity of the particle is the derivative of the position function with respect to time, i.e., \[ v(t) = \frac{ds}{dt} \] Given: \[ s(t) = \frac{2}{\sqrt{5t + 6}} \]

Step 2 - Differentiate the position function

Apply the chain rule to find \( \frac{ds}{dt} \). Let \[ u = 5t + 6 \] so that \[ s(t) = \frac{2}{\sqrt{u}} \]. Then, differentiate with respect to \(u\): \[ \frac{ds}{du} = \frac{d}{du}\left(\frac{2}{\sqrt{u}}\right) = \frac{d}{du}(2u^{-1/2}) = 2 \cdot \left(-\frac{1}{2}\right) u^{-3/2} = -\frac{1}{u^{3/2}} \] Next, differentiate \( u = 5t + 6 \) with respect to \(t\): \[ \frac{du}{dt} = 5 \] Then, \[ \frac{ds}{dt} = \frac{ds}{du} \cdot \frac{du}{dt} \] So, \[ v(t) = -\frac{1}{(5t + 6)^{3/2}} \cdot 5 = \frac{-5}{(5t + 6)^{3/2}} \]

Step 3 - Substitute the value of \( t_1 \)

Now, substitute \(t_1 = 2\) into the velocity equation: \[ v(2) = \frac{-5}{(5(2) + 6)^{3/2}} = \frac{-5}{(10 + 6)^{3/2}} = \frac{-5}{16^{3/2}} \] Since \( 16 = 4^2 \), we get: \[ 16^{3/2} = (4^2)^{3/2} = 4^3 = 64 \] Therefore, \[ v(2) = \frac{-5}{64} \]

Key Concepts

position function derivative

To understand the concept of instantaneous velocity, we need to first differentiate the position function. The position function, denoted as \(s(t)\), describes the location of a particle at any given time \(t\). When we differentiate this function with respect to time, we obtain the velocity function \(v(t)\). This derivative essentially tells us how fast the position of the particle is changing at any moment.
In the given exercise, the position function is \(s(t) = \frac{2}{\sqrt{5t + 6}}\). By differentiating this position function using calculus, we can then find the instantaneous velocity.

chain rule in differentiation

To find the derivative of a composite function, we use the chain rule. This rule is crucial when dealing with the differentiation of functions nested inside other functions.
For example, if we have a function \(s(t) = \frac{2}{\sqrt{5t + 6}}\), we can let \(u = 5t + 6\) and rewrite the function in terms of \(u\) as \(s(u) = \frac{2}{\sqrt{u}}\).
Next, we apply the chain rule: differentiate \(s\) with respect to \(u\) to get \(\frac{ds}{du}\), and then differentiate \(u\) with respect to \(t\) to get \(\frac{du}{dt}\). Finally, we multiply these derivatives together to get the overall derivative: \(\frac{ds}{dt} = \frac{ds}{du} \cdot \frac{du}{dt}\).
In the exercise, we found:

• \( \frac{ds}{du} = -\frac{1}{u^{3/2}} \)
• \( \frac{du}{dt} = 5 \)

So the velocity function becomes \( v(t) = \frac{-5}{(5t + 6)^{3/2}} \).

evaluating derivatives at specific points

Once the derivative (instantaneous velocity function) is found, the next step is to evaluate it at specific points to get the instantaneous velocity at those times.
In the exercise, we're asked to find the velocity at \(t_1 = 2\) seconds. We substitute \(t = 2\) into the velocity function \(v(t) = \frac{-5}{(5t + 6)^{3/2}}\). Doing this, we get:

• \( v(2) = \frac{-5}{(5(2) + 6)^{3/2}} = \frac{-5}{16^{3/2}} \)

We then simplify further: \( 16^{3/2} = (4^2)^{3/2} = 4^3 = 64 \).
Thus, \( v(2) = \frac{-5}{64} \).

particle motion in calculus

Calculus is fundamental for understanding the motion of particles along a path. The position function \(s(t)\) describes how the location of a particle changes over time.
By differentiating \(s(t)\), we obtain the velocity function \(v(t)\), which tells us the rate of change of the particle's position – its speed and direction at any moment.
The instantaneous velocity, derived from the position function, answers 'how fast?' and 'in which direction?' the particle is moving at a specific time.
In the provided exercise, we determined the particle's velocity as \( v(t) = \frac{-5}{(5t + 6)^{3/2}} \), which tells us the particle’s motion dynamics depending on the given time \(t\).
For deeper insights, acceleration, the next derivative of velocity, can also be derived to understand how the particle's velocity changes over time. But for this scenario, we focused on velocity, giving us the particle's speed at precise moments, such as at \(t_1 = 2\) seconds.