Problem 41 - Calculus III

Question

Locate the highest point on the curve \(\mathbf{r}(t)=\left\langle 6 t, 6 t-t^{2}\right\rangle\) and give the value of the function at this point.

Step-by-Step Solution

Answer

The highest point is at \( (18, 9) \) with a maximum function value of \( 9 \).

Step 1: Identify the Curve Components

The curve is given by the vector function \( \mathbf{r}(t)=\left\langle 6t, 6t-t^{2} \right\rangle \), which represents a parametric equation in terms of \( t \), where \( x(t) = 6t \) and \( y(t) = 6t - t^2 \).

Step 2: Expression for y(t)

The function we are interested in maximizing is \( y(t) = 6t - t^2 \). We can also interpret this as the height of the curve at any point \( t \). We need to find the maximum value of this function.

Step 3: Differentiate y(t)

To find the maximum value of \( y(t) \), differentiate it with respect to \( t \). This gives us \( y'(t) = \frac{d}{dt}(6t - t^2) = 6 - 2t \).

Step 4: Set Derivative to Zero

To locate the extrema, solve \( y'(t) = 0 \). This gives us the equation \( 6 - 2t = 0 \). Solving for \( t \), we find \( t = 3 \).

Step 5: Second Derivative Test

Check the nature of the stationary point by finding \( y''(t) \). Differentiate \( y'(t) \) to get \( y''(t) = -2 \), which is negative, indicating a maximum at \( t = 3 \).

Step 6: Calculate Maximum Height

Substitute \( t = 3 \) back into \( y(t) \) to calculate the height: \( y(3) = 6(3) - 3^2 = 18 - 9 = 9 \).

Step 7: Determine x-coordinate at t = 3

Substitute \( t = 3 \) into \( x(t) = 6t \) to find the x-coordinate: \( x(3) = 6(3) = 18 \).

Step 8: Confirm the Highest Point

Thus, the highest point on the curve occurs at the coordinates \((18, 9)\). The value of the function at this point is \( 9 \).

Key Concepts

Parametric Equations

Parametric equations allow us to express a curve using a parameter, typically denoted as \( t \), rather than a single independent variable like \( x \) or \( y \). This can be useful for representing more complex curves that may not be easily described with standard functions. In our exercise, the parametric equations \( x(t) = 6t \) and \( y(t) = 6t - t^2 \) describe the path of a curve in the plane.
Understanding parametric equations involves evaluating each component separately:

• \( x(t) = 6t \): This describes the horizontal motion and shows that as \( t \) increases, \( x \) increases linearly.
• \( y(t) = 6t - t^2 \): This describes vertical motion, and it's a quadratic function that forms a parabola when plotted against \( t \).

By combining these two components, we obtain the trajectory of the curve in two-dimensional space. This is crucial for analyzing motion along a path when both position and direction change with time.

Differentiation

Differentiation is a core concept in calculus used to determine the rate at which a function changes. In the context of parametric equations, we focus on differentiating the \( y(t) \) component because we aim to find where the curve reaches its highest point. For this problem, the function is \( y(t) = 6t - t^2 \).
The derivative \( y'(t) \) represents the rate of change of \( y \) concerning \( t \), and it's calculated as follows:

• Find \( y'(t) = \frac{d}{dt}(6t - t^2) = 6 - 2t \).

This derivative tells us how the height of the curve changes as \( t \) changes. By setting \( y'(t) \) to zero, we identify the points where the curve's height does not change, indicating potential maximum or minimum points.

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function, which are often critical points in applications. In our exercise, we seek to maximize \( y(t) = 6t - t^2 \), essentially finding the highest point on the curve.
To locate these critical points, we set the derivative \( y'(t) \) to zero:

• Solve \( 6 - 2t = 0 \) to find \( t = 3 \).

With \( t = 3 \), we have a candidate for the maximum height. Substituting this value into \( y(t) \) verifies it:

• Calculate \( y(3) = 6(3) - 3^2 = 9 \).

Thus, the maximum height achieved is 9 units at \( t = 3 \). This approach ensures we use calculus tools effectively to solve practical problems involving optimum values.

Second Derivative Test

The Second Derivative Test uses the second derivative of a function to ascertain the nature of its critical points. For our curve, this involves the function \( y(t) = 6t - t^2 \) and its derivatives. Once we find the first derivative \( y'(t) = 6 - 2t \), we compute its second derivative to apply this test:

• The second derivative is \( y''(t) = -2 \).

A negative second derivative, as in this case, indicates that the critical point is a local maximum. Thus, the point \( t = 3 \) indeed corresponds to a peak of the curve.
This test is a powerful tool as it not only confirms the location of extreme values but also ensures we correctly classify these as either maxima or minima, helping us interpret the behavior of the function accurately.