Question

PROBLEM SOLVING The engine torque y (in foot-pounds) of one model of car is given by y = ?3.75x2 + 23.2x + 38.8, where x is the speed (in thousands of revolutions per minute) of the engine. a. Find the engine speed that maximizes torque. What is the maximum torque? b. Explain what happens to the engine torque as the speed of the engine increases.

Short answer

The engine speed that maximizes torque is approximately 3093 rpm and the maximum torque at this speed is approximately 45.847 foot-pounds. The torque increases with speed until it reaches a maximum at 3093 rpm, after which it decreases.

Step-by-step solution

Differentiate the torque function

The torque function y is given as \(y = -3.75x^2 + 23.2x + 38.8\). To find the engine speed that maximizes the torque function, we need to find the critical points of the function. This is done by taking the derivative of the function and setting it equal to zero. The derivative of y with respect to x is \((dy/dx) = 2*(-3.75)x + 23.2\). Now, we set this equal to zero for solving it.

Solve to Find Critical Point

Set 2*(-3.75)x + 23.2 =0. From here we can solve for x as \((23.2)/(2*3.75)\), the negative solution is disregarded as speed cannot be negative. This yields x =3.093. Thus at engine speed of approximately 3093 revolutions per minute, maximum torque is possible.

Compute the Maximum Torque

To find the maximum torque, substitute x=3.093 into the original equation for y. Hence, \(y = -3.75*3.093^2 + 23.2*3.093 + 38.8\), which comes to about 45.847 foot-pounds.

Interpret the Torque Behaviour

Judging by the original equation which is a standard quadratic equation \(y = ax^2 + bx + c\), where the coefficient of \(x^2\), 'a', is negative. This suggests that the graph of the function opens downwards, indicating that the torque increases as the speed increases until a maximum is reached at 3093 rpm, after which the torque decreases.

Key concepts

Understanding Calculus Applications

In this exercise, we explore how calculus can be applied to optimize certain parameters, in this case, engine torque. Using calculus, specifically differentiation, we can find the maximum or minimum values of functions that model real-world situations. This is crucial in engineering and design processes, such as optimizing engine performance. Here, we're dealing with a quadratic function which is often used to model phenomena that show an initial increase, followed by a decrease. In our scenario, understanding the torque curve's behavior, when plotted against engine speed, helps us pinpoint the optimal speed for maximum efficiency. The principles of calculus enable us to analyze the function's graph and make precise predictions about its behavior.

The Role of Differentiation

Differentiation is a core concept in calculus and crucial for finding critical points where functions reach their extreme values, such as maxima or minima. By differentiating, we can compute the rate of change of the function with respect to its variable, helping us understand how the function behaves as its input changes.
For this specific quadratic function, \[ y = -3.75x^2 + 23.2x + 38.8 \] we computed the derivative, \[ \frac{dy}{dx} = -7.5x + 23.2 \] The derivative evaluates how the torque (output) changes as the engine speed (input) increases. Setting the derivative equal to zero \[ -7.5x + 23.2 = 0 \] allows us to solve for critical points, indicating where the maximum torque occurs. This tells us the point (at 3093 revolutions per minute) where the engine performs optimally.Differentiation thus acts as an analytical tool in optimizing and predicting performance in various systems.

Critical Points and Their Importance

Critical points are values of \( x \) where the derivative of a function is zero or undefined. They are essential for determining where a function's slope is zero, indicating possible maxima or minima points. These points are critical for identifying optimal performance in systems like engines, allowing designers to enhance efficiency.

• In the problem, we used the derivative \( \frac{dy}{dx} = -7.5x + 23.2 \) to find critical points.
• Solving \( -7.5x + 23.2 = 0 \) gave us \( x = 3.093 \), suggesting 3093 rpm results in the maximum torque.
• By substituting \( x = 3.093 \) back into the original function, we calculated the maximal torque value.

Understanding critical points is vital for designing systems that operate at maximum efficiency. They reveal the limits and potential of systems governed by mathematical functions, showcasing the power of calculus in real-world applications.