Problem 5 - Calculus of a Single Variable

Question

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=2 x^{2}+1 ; \frac{d x}{d t}=2 \text { centimeters per second }} \\ {\begin{array}{ll}{\text { (a) } x=-1} & {\text { (b) } x=0} & {\text { (c) } x=1}\end{array}}\end{array} $$

Step-by-Step Solution

Answer

The rate of change \(dy/dt\) when \(x=-1\) is -8 cm/s, when \(x=0\) is 0 cm/s, and when \(x=1\) is 8 cm/s.

Step 1: Calculate the derivative of y with respect to x

Take the derivative of the function \(y = 2x^2 + 1\) with respect to \(x\). This yields \(dy/dx = 4x\).

Step 2: Apply the chain rule for each scenario

Use the chain rule (\(dy/dt = dy/dx . dx/dt\)) to compute \(dy/dt\) for each \(x\) value. Since \(dx/dt\) is the same (2 cm/s) for all scenarios, only replace \(x\) in \(dy/dx = 4x\) with the given value, and then multiply with \(dx/dt\).

Step 3: Compute the rate of change for x = -1

When \(x = -1\), the rate of change \(dy/dt\) would be \(4*(-1)*2 = -8\) cm/s.

Step 4: Compute the rate of change for x = 0

When \(x = 0\), the rate of change \(dy/dt\) would be \(4*0*2 = 0\) cm/s.

Step 5: Compute the rate of change for x = 1

When \(x = 1\), the rate of change \(dy/dt\) would be \(4*1*2 = 8\) cm/s.

Key Concepts

Derivative

A derivative, in the simplest terms, tells us how a quantity changes with respect to another. Imagine you are driving a car and want to know how fast you are going at a specific moment, not just your average speed. That's what derivatives do for us in math! They give us the instantaneous rate of change.

For the given function, which was **\( y = 2x^2 + 1 \)**, we seek to determine how \(y\) changes as \(x\) changes.

• We "take the derivative" of \(y\), which means we find a new expression, \(dy/dx\), representing the rate at which \(y\) changes with respect to \(x\).
• By differentiating \(y = 2x^2 + 1\), we find \(dy/dx = 4x\).

This result shows how \(y\) changes per unit change in \(x\), helping us understand the behavior of the function at different points.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to handle situations where one variable depends on another, which in turn depends on a third variable. It lets us track the variation through these dependencies step-by-step, much like following a chain of reasoning.

• In our example, \(y\) depends on \(x\), and \(x\) changes over time \(t\) with a known rate \(dx/dt\).
• The chain rule formula is: \( dy/dt = (dy/dx) \cdot (dx/dt) \).

This means the rate of change of \(y\) with respect to time can be determined by multiplying the rate of \(y\) with \(x\) by the rate of \(x\) with respect to \(t\). By substituting \(dy/dx = 4x\) and \(dx/dt = 2\) cm/s into this formula, we can find the time rate of change for different \(x\) values.

Rate of Change

The rate of change can be thought of as how fast one quantity changes, usually over time. In our exercise, it is specifically about how fast \(y\) changes as \(x\) changes over time.

• For \(x = -1\), the calculated \(dy/dt\) was \(-8\) cm/s, indicating that \(y\) is decreasing because the negative sign shows a downward change.
• When \(x = 0\), \(dy/dt\) equaled \(0\), signaling no change in \(y\) at that instant since the rate is zero.
• For \(x = 1\), \(dy/dt\) was \(8\) cm/s, showing that \(y\) is increasing at this point.

Such calculations of the rate of change are crucial in many fields, such as physics, to predict behaviors and understand underlying trends in the movement or growth of a system.