Problem 87 - Calculus and its applications

Question

Graph \(s,\) v, and a over the given interval. Then use the graphs to determine the point(s) at which the velocity switches from increasing to decreasing or from decreasing to increasing. $$ s(t)=t^{4}+t^{3}-4 t^{2}-2 t+4 ; \quad[-3,3] $$

Step-by-Step Solution

Answer

The velocity changes behavior at approximate critical points given by solving \(12t^2 + 6t - 8 = 0\).

Step 1: Find the Derivative for Velocity

To graph velocity, we need to find the derivative of the position function \(s(t)\). The velocity function \(v(t)\) is the first derivative of \(s(t)\): \[v(t) = s'(t) = \frac{d}{dt}(t^4 + t^3 - 4t^2 - 2t + 4) = 4t^3 + 3t^2 - 8t - 2.\]

Step 2: Find the Second Derivative for Acceleration

To graph acceleration, we find the derivative of the velocity function \(v(t)\). The acceleration function \(a(t)\) is the second derivative of \(s(t)\):\[a(t) = v'(t) = \frac{d}{dt}(4t^3 + 3t^2 - 8t - 2) = 12t^2 + 6t - 8.\]

Step 3: Graph the Position, Velocity, and Acceleration

Now, graph each of the functions \(s(t)\), \(v(t)\), and \(a(t)\) over the interval \([-3, 3]\). This can be done using graphing software or by plotting key points and considering the general shape of polynomial graphs. Note where each graph intersects and changes direction.

Step 4: Determine Critical Points of Velocity

The points where velocity switches from increasing to decreasing, or vice versa, are critical points. These occur where the acceleration, or the derivative of velocity \(a(t)\), is zero:\[a(t) = 12t^2 + 6t - 8 = 0.\]Solve this quadratic equation to find the critical points.

Step 5: Solve for Critical Points

Solve the equation \(12t^2 + 6t - 8 = 0\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 12\), \(b = 6\), and \(c = -8\):\[t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 12 \cdot (-8)}}{2 \cdot 12}.\]Calculate to find the approximate values of \(t\).

Step 6: Calculate and Analyze the Roots

Calculate the roots:\[b^2 - 4ac = 36 + 384 = 420\]\[t = \frac{-6 \pm \sqrt{420}}{24}.\]Calculate the approximate values of \(t\), which are the critical points where velocity switches behavior. This results in approximate values, which can be plotted on the velocity graph.

Step 7: Conclusion from Graphs

By examining the graphs, the points where the sign of acceleration changes (indicating a change in velocity behavior from increasing to decreasing or vice versa) are roughly at the values computed. These are the points where the velocity transition occurs.

Key Concepts

Derivative

A derivative in calculus helps us understand how functions change. Specifically, it shows the rate at which a function's value shifts as its input changes. When we think about a position function like \(s(t)\), the derivative tells us about the object's velocity, or how fast the position is changing at any given moment.

To find the velocity function \(v(t)\), we calculate the first derivative of the position function:

• For \(s(t) = t^4 + t^3 - 4t^2 - 2t + 4\), the derivative is \(v(t) = 4t^3 + 3t^2 - 8t - 2\).

This process involves using power rules, where you multiply the exponent by the coefficient and then subtract one from the exponent.

In our exercise, this derivative is crucial because it allows us to graph the velocity function and look for critical points where velocity changes behavior.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations. These are equations that take the form \(ax^2 + bx + c = 0\). The formula allows us to find the values of \(x\) that satisfy the equation. It is expressed as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

To solve the acceleration function \(a(t) = 12t^2 + 6t - 8 = 0\), we substitute \(a = 12\), \(b = 6\), and \(c = -8\) into this formula.

Doing so reveals the critical points where the behavior of the velocity (increase or decrease) changes. These critical points are essential for understanding how and when the object described in our function changes speed.

Acceleration

Acceleration is the rate of change of velocity over time. Just as the velocity is the first derivative of position, acceleration is the derivative of velocity. It tells us how quickly or slowly an object's velocity is changing.

For our function, the acceleration \(a(t)\) is found by taking the derivative of the velocity function:

• \(v(t) = 4t^3 + 3t^2 - 8t - 2\)
• \(a(t) = 12t^2 + 6t - 8\)

Finding when \(a(t) = 0\) helps identify the critical points where the velocity's rate of change transitions. This can indicate a change from speeding up to slowing down or vice versa. By analyzing these zero points, we better understand the object's motion and can graph it to visualize shifts in speed.

Polynomial Graphs

Polynomial graphs are graphs of polynomial functions, which are expressions involving sums of powers of a variable. These graphs can take on various shapes, depending on the degree (highest power) and the coefficients of the polynomial.

The main graph types involved in our exercise are:

• Position Function Graph \(s(t)\) - fourth degree:
• Velocity Function Graph \(v(t)\) - third degree:
• Acceleration Function Graph \(a(t)\) - second degree (quadratic):

When plotting these functions in the interval \([-3, 3]\), we observe key behaviors like turning points and intercepts that guide us in interpreting physical behaviors such as velocity and acceleration changes. Understanding how each polynomial is shaped gives insight into the motion described, enhancing comprehension of calculus concepts through visualization.