Problem 24 - Calculus

Question

If \(f(t)=\frac{1}{2} a t^{2}+b t+c,\) what is \(v(t) ?\) What is the slope of \(v(t) ?\) When does \(f(t)\) equal \(41,\) if \(a=b=c=1 ?\)

Step-by-Step Solution

Answer

v(t) = at + b\,; slope = a; \(f(t) = 41\) when \(t = 8\).

Step 1: Find v(t) by differentiating f(t)

The function given is \[ f(t) = \frac{1}{2}at^2 + bt + c. \]To find the velocity function \(v(t)\) with respect to time \(t\), you need to differentiate \(f(t)\) with respect to \(t\). This gives:\[ v(t) = \frac{d}{dt}(f(t)) = \frac{d}{dt}\left(\frac{1}{2}at^2 + bt + c\right). \] Differentiating each term separately, we get:\[ v(t) = at + b. \]

Step 2: Determine the slope of v(t)

To determine the slope of \(v(t)\), differentiate \(v(t) = at + b\) with respect to \(t\). This gives us:\[ \frac{d}{dt}(v(t)) = \frac{d}{dt}(at + b) = a. \]Thus, the slope of \(v(t)\) is \(a\).

Step 3: Substitute values to find when f(t) equals 41

We need to find the time \(t\) when \(f(t) = 41\), given \(a = b = c = 1\). Substitute these values into the expression for \(f(t)\):\[ f(t) = \frac{1}{2}(1)t^2 + (1)t + (1). \]Simplifying, we have:\[ f(t) = \frac{1}{2}t^2 + t + 1. \]Set \(f(t) = 41\), so:\[ \frac{1}{2}t^2 + t + 1 = 41. \]Rearrange the equation:\[ \frac{1}{2}t^2 + t - 40 = 0. \]Multiply through by 2 to clear the fraction:\[ t^2 + 2t - 80 = 0. \]

Step 4: Solve the quadratic equation for t

Use the quadratic formula to solve:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \(a = 1\), \(b = 2\), and \(c = -80\).Substituting, we get:\[ t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-80)}}{2 \cdot 1}. \]Simplifying further:\[ t = \frac{-2 \pm \sqrt{4 + 320}}{2}. \]\[ t = \frac{-2 \pm \sqrt{324}}{2}. \]\[ t = \frac{-2 \pm 18}{2}. \]The solutions are:\[ t = \frac{16}{2} = 8 \quad \text{and} \quad t = \frac{-20}{2} = -10. \]Hence, \(f(t) = 41\) when \(t = 8\). Negative \(t\) is not considered as it usually represents non-physical solutions like negative time.

Key Concepts

Differentiation

Differentiation is a fundamental concept in differential calculus, and it refers to the process of finding the derivative of a function. A derivative represents the rate at which a quantity changes with respect to another. In simpler terms, differentiation helps us determine how a function behaves as its input, usually time or space, changes.

In the context of the function given, which is a quadratic function of form:

• \( f(t) = \frac{1}{2}at^2 + bt + c \), where \(a\), \(b\), and \(c\) are constants, the derivative \(v(t)\) is found by differentiating \(f(t)\) with respect to the variable \(t\).
• Differentiating each component separately, the power rule is used to find: \( \frac{d}{dt}(\frac{1}{2}at^2 + bt + c) = at + b \).

This derivative, \(v(t)\), tells us the instantaneous rate of change of the function \(f(t)\) with respect to time \(t\). In physical terms, when \(f(t)\) models position, \(v(t)\) represents velocity.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They are commonly expressed in the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are coefficients. These equations are pivotal in various scientific and engineering domains as they describe phenomena such as parabolic motion and acceleration.

In the problem at hand, after substituting the values \(a = b = c = 1\), the equation becomes a simplified quadratic equation

• \( \frac{1}{2}t^2 + t + 1 = 41 \).
• To solve this, it needs to be rearranged to standard form: \( \frac{1}{2}t^2 + t - 40 = 0 \).
• Multiplying through by 2 to eliminate the fraction, we have \( t^2 + 2t - 80 = 0 \).

The solutions to this quadratic equation are found using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), yielding the roots \( t = 8 \) and a non-physical negative root \( t = -10 \). These solutions tell us when the function \( f(t) \) equals 41.

Velocity Function

The velocity function, denoted as \(v(t)\), is the derivative of the position function to time \(t\). It indicates how fast an object is moving and in which direction. Essentially, if you have a position-time function like \( f(t) \), differentiating it gives you the velocity function.

In this exercise, we start with

• \( f(t) = \frac{1}{2}at^2 + bt + c \), representing position over time.
• Differentiation gives \( v(t) = at + b \).

This outcome signifies that velocity is linearly dependent on time for the given quadratic position function. It also highlights that the velocity function is defined by the linear component coefficients of the original function.

Slope of a Function

The slope of a function refers to the rate at which the function's value changes as its input changes. In calculus, the slope of a function at any point can be found by differentiating it. This is particularly meaningful for linear functions, like the velocity function, where the slope is constant.

For this problem, the velocity function \(v(t) = at + b\) has a slope described by the constant \(a\).

• By differentiating \(v(t)\), you end up with the slope: \( \frac{d}{dt}(at + b) = a \).
• A constant slope indicates that the velocity changes at a uniform rate over time.

The significance of the slope here is that it provides direct insight into how quickly the velocity is changing, effectively showing a constant acceleration in the context of motion modeled by the initial quadratic function \(f(t)\).