Problem 248 - Calculus Volume 1

Question

[T] Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. The best quadratic fit to the data is \(q(t)=5 x^{2}-11 x+49,\) shown in the accompanying graph. Integrate \(q\) to estimate the total distance driven over the 3 hours.

Step-by-Step Solution

Answer

The total distance driven over 3 hours is 142.5 units.

Step 1: Understand the Problem

We need to estimate the total distance driven over 3 hours by integrating the quadratic function \( q(t) = 5t^2 - 11t + 49 \). The integration of this function over the interval from 0 to 3 will give us the total distance traveled.

Step 2: Set Up the Integral

To find the total distance traveled, we need to compute the definite integral of the quadratic function from 0 to 3. This is expressed as \( \int_{0}^{3} (5t^2 - 11t + 49) \, dt \).

Step 3: Integrate the Function

We calculate the integral by finding the antiderivative of \(q(t)\). The antiderivative of \(5t^2 \) is \( \frac{5}{3}t^3 \), the antiderivative of \(-11t\) is \( -\frac{11}{2}t^2 \), and the antiderivative of \(49\) is \(49t\). So, the antiderivative of \(q(t)\) is \(\frac{5}{3}t^3 - \frac{11}{2}t^2 + 49t\).

Step 4: Evaluate the Integral at the Bounds

Evaluate the antiderivative at the upper bound and the lower bound. Calculate \(\frac{5}{3}(3^3) - \frac{11}{2}(3^2) + 49(3)\) and subtract \( \frac{5}{3}(0^3) - \frac{11}{2}(0^2) + 49(0) \).

Step 5: Simplify the Evaluation

Calculate each term at \(t = 3\):- \(\frac{5}{3}(27) = 45\)- \(\frac{11}{2}(9) = 49.5\)- \(49(3) = 147\)Thus, the result at \(t = 3\) is \(45 - 49.5 + 147 = 142.5\). Since the result at \(t = 0\) is 0, the total distance traveled is 142.5.

Key Concepts

Quadratic Functions

Quadratic functions are polynomials of degree two. They have the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These functions create a parabola when graphed, either opening upwards if \( a > 0 \) or downwards if \( a < 0 \). In the context of a road trip, a quadratic function can represent the speed over time, helping us predict travel trends.

For our problem, the quadratic formula is given as \( q(t) = 5t^2 - 11t + 49 \). This function models the speed at different times during a 3-hour trip, allowing us to use calculus to compute the total distance covered.

Definite Integral

The definite integral is a fundamental concept in calculus with applications in finding areas, volumes, and accumulated quantities. Denoted as \( \int_{a}^{b} f(x) \, dx \), it represents the area under the curve \( f(x) \) from \( x = a \) to \( x = b \).

In our exercise, we're evaluating the definite integral of the quadratic function \( q(t) = 5t^2 - 11t + 49 \) from 0 to 3. The result will give us the total distance traveled, as the area under the speed-time graph equates to distance.

Antiderivative

Finding an antiderivative involves determining a function whose derivative is the given function. For polynomials, the antiderivative can be found by reversing the power rule of differentiation. Each term of the polynomial is treated separately through this process.

In steps, for the function \( q(t) = 5t^2 - 11t + 49 \):

• The antiderivative of \( 5t^2 \) is \( \frac{5}{3}t^3 \)
• The antiderivative of \( -11t \) is \( -\frac{11}{2}t^2 \)
• The antiderivative of \( 49 \) is \( 49t \)

By adding these, the complete antiderivative becomes \( \frac{5}{3}t^3 - \frac{11}{2}t^2 + 49t \). This helps in evaluating the definite integral required for calculating total distance.

Distance Estimation

Distance estimation in integral calculus involves finding the total distance traveled by a moving object when a speed-time function is provided. Integrating the speed function over a defined interval gives an accurate estimation of distance covered.

For our road trip exercise, by integrating the speed function \( q(t) = 5t^2 - 11t + 49 \) from \( t = 0 \) to \( t = 3 \):

• Calculate \( \frac{5}{3}(3^3) = 45 \)
• Subtract \( \frac{11}{2}(3^2) = 49.5 \)
• Add \( 49(3) = 147 \)

The final calculation results in \( 142.5 \), giving us the estimated total distance traveled over the 3 hours of the road trip as 142.5 units.