Problem 61 - Technical Mathematics with Calculus

Question

The position \(s\) of a point in a mechanism is given by $$s=5 t^{2}+3 t \text { in. }$$ where \(t\) is the time in seconds. The velocity of the point is found by taking the first derivative of the displacement, or \(d s / d t .\) Take the derivative and evaluate it at \(t=3.55 \mathrm{s}\).

Step-by-Step Solution

Answer

The velocity of the point at \(t = 3.55\) seconds is \(38.5\) inches per second.

Step 1: Identify the function to differentiate

Assess the given position function with respect to time, which is in the form of a polynomial. This function is given as \(s = 5t^2 + 3t\) inches.

Step 2: Differentiate the position function

Differentiate the position function \(s(t)\) with respect to time \(t\) to find the velocity function, using the power rule. Taking the derivative of each term in \(s(t)\), the velocity function will be \(v(t) = \frac{ds}{dt} = \frac{d}{dt}(5t^2) + \frac{d}{dt}(3t)\).

Step 3: Apply the power rule for each term

Apply the power rule, which states that the derivative of \(t^n\) with respect to \(t\) is \(n \cdot t^{n-1}\). So, for the first term \(5t^2\), it becomes \(2 \cdot 5t^{2-1} = 10t\), and for the second term \(3t\), it becomes \(3\).

Step 4: Combine the derivatives

Combine the derivatives of both terms of the position function to get the velocity function: \(v(t) = 10t + 3\).

Step 5: Evaluate the velocity at a specific time

Substitute \(t = 3.55\) seconds into the velocity function to find the velocity at that time: \(v(3.55) = 10(3.55) + 3\).

Step 6: Calculate the numerical value

Perform the multiplication and addition to find the value of velocity at \(t = 3.55\) seconds: \(v(3.55) = 10(3.55) + 3 = 35.5 + 3 = 38.5\) inches per second.

Key Concepts

Derivative of Position Function

In kinematics, the derivative of the position function with respect to time gives us the velocity of an object. This fundamental concept is grounded in calculus and is essential for understanding motion. When you are given a position function like \(s = 5t^2 + 3t\), the expression represents the distance traveled by an object along a straight line as a function of time.

To find the rate of change of position—or simply put, the velocity—we calculate the derivative of this function. This involves taking the derivative of each term in the position function with respect to time. For students grappling with this concept, it's important to view the derivative as a rate: how quickly the position is changing at any given instant of time.

Power Rule Differentiation

One of the simplest and most widely used techniques for differentiating polynomial functions is power rule differentiation. It is a straightforward process: when you have a term in the form \(ct^n\), where \(c\) is a constant and \(n\) is a power, its derivative is \(n \times ct^{n-1}\).

In the context of the exercise, the power rule is applied to the terms \(5t^2\) and \(3t\), yielding \(10t\) and \(3\) respectively.

Remember the Power Rule Shortcut:

To make differentiation faster, remember that for any term \(t^n\), simply multiply the exponent by the coefficient and then decrease the exponent by one.

Velocity Function Evaluation

After using the power rule to find the velocity function, which is the derivative of the position function, the next step is to evaluate this velocity function at a specific time. This evaluation gives us the velocity of the object at that particular instant.

In the presented problem, we substitute \(t = 3.55\) seconds into \(v(t) = 10t + 3\) to find the instantaneous velocity. When evaluating, careful substitution and arithmetic are key—this ensures accuracy when computing the object's velocity. The result is a concrete figure that represents how fast the object is moving in inches per second at the 3.55-second mark.