Problem 43 - Basic Technical Mathematics with Calculus

Question

Solve the given applied problem. A computer analysis of the power \(P\) (in W) used by a pressing machine shows that \(P=50 i-3 i^{2},\) where \(i\) is the current (in A). Sketch the graph of \(P=f(i)\).

Step-by-Step Solution

Answer

Sketch the parabola opening downwards, with intercepts at \(i = 0\) and \(i = \frac{50}{3}\), and vertex at \(\left(\frac{25}{3}, -\frac{625}{3}\right)\).

Step 1: Understand the Function

The given function is a quadratic function in the form \(P = f(i) = 50i - 3i^2\). This represents a parabola opening downwards because the coefficient of \(i^2\) is negative.

Step 2: Identify Key Characteristics

To sketch the graph, identify important characteristics such as the vertex, axis of symmetry, and intercepts. The standard form of a quadratic is \(ax^2 + bx + c\). Here, \(a = -3\), \(b = 50\), and \(c = 0\). The axis of symmetry is given by \(i = -\frac{b}{2a}\).

Step 3: Calculate the Vertex

The axis of symmetry is \(i = -\frac{50}{2(-3)} = \frac{50}{6} = \frac{25}{3}\). Substitute \(i = \frac{25}{3}\) back into the function to find the maximum power: \(P = 50 \times \frac{25}{3} - 3 \times \left(\frac{25}{3}\right)^2 = \frac{1250}{3} - \frac{1875}{3} = -\frac{625}{3}\). Thus, the vertex is \(\left(\frac{25}{3}, -\frac{625}{3}\right)\).

Step 4: Calculate Intercepts

To find the i-intercepts, set \(P = 0\): \(50i - 3i^2 = 0\). Factor to get \(i(50 - 3i) = 0\). Thus, the intercepts are \(i = 0\) and \(i = \frac{50}{3}\).

Step 5: Sketch the Graph

On a graph, plot the vertex at \(\left(\frac{25}{3}, -\frac{625}{3}\right)\) and the intercepts at \(i = 0\) and \(i = \frac{50}{3}\). Draw a parabola opening downwards, with the axis of symmetry at \(i = \frac{25}{3}\). The graph should start at the intercepts, peak at the vertex, and symmetrically mirror the other side.

Step 6: Final Review and Adjustments

Ensure that the essential points (vertex, intercepts) are plotted accurately, and the graph properly reflects the downward-opening nature due to the negative coefficient.

Key Concepts

Vertex of a Parabola

In quadratic functions, the vertex of a parabola is a crucial point. It represents the peak or the lowest point of the parabola depending on the orientation. In our problem, the quadratic function given is \(P = 50i - 3i^2\). This parabola opens downwards since the coefficient of \(i^2\) is negative.

The vertex is the point where the parabola changes direction. For functions of the form \(ax^2 + bx + c\), the vertex can be found using the formula:

• The \(i\)-coordinate of the vertex: \(i = -\frac{b}{2a}\)
• The \(P\)-coordinate is found by substituting \(i\) back into the function.

In our case:

• \(i = -\frac{50}{2(-3)} = \frac{25}{3}\)
• \(P = f(\frac{25}{3}) = 50 \times \frac{25}{3} - 3 \times (\frac{25}{3})^2\)
• Thus, the vertex is \(\left(\frac{25}{3}, -\frac{625}{3}\right)\).

Understanding where the vertex is located helps in predicting the behavior of the parabola when plotted.

Axis of Symmetry

The axis of symmetry in a parabola is an imaginary vertical line that passes through the vertex, dividing the parabola into two mirror images. This is essential for understanding how the graph will look.

For any quadratic function \(ax^2 + bx + c\), the axis of symmetry can be calculated with the formula \(i = -\frac{b}{2a}\). It's the same value used in finding the \(i\)-coordinate of the vertex.

In our specific function \(P = 50i - 3i^2\):

• We find \(i = -\frac{50}{2(-3)} = \frac{25}{3}\).
• So, the axis of symmetry is the line \(i = \frac{25}{3}\).

This line is crucial as it aids in creating a symmetrical graph, ensuring both sides of the parabola reflect across it. While sketching, having this line drawn helps set the foundational structure for the curve.

Graphing Quadratics

Graphing a quadratic function involves plotting key points and understanding its general shape. For the function \(P = 50i - 3i^2\), it's necessary to determine:

• The vertex, \(\left(\frac{25}{3}, -\frac{625}{3}\right)\)
• The axis of symmetry, \(i = \frac{25}{3}\)
• Intercepts where the parabola crosses the axis.

Find the \(i\)-intercepts by setting the equation to zero: \(50i - 3i^2 = 0\). Solving gives:

• \(i = 0\)
• \(i = \frac{50}{3}\)

With these points, you can start sketching the graph:

• Plot the vertex and intercepts on a coordinate plane.
• Draw a smooth curve opening downward from the vertex through the intercepts, ensuring to mirror the curve along the axis of symmetry.

Properly identifying and plotting these elements ensures a clear and accurate representation of the quadratic function.