Question

In Exercises, use the given values to find \(d y / d t\) and \(d x / d t\) $$ \begin{aligned} &y=2\left(x^{2}-3 x\right)\\\ &\text { (a) } \frac{d y}{d t} \quad x=3, \frac{d x}{d t}=2\\\ &\text { (b) } \frac{d x}{d t} \quad x=1, \frac{d y}{d t}=5 \end{aligned} $$

Short answer

a) \(dy/dt = 12\), b) \(dx/dt = -2.5\).

Step-by-step solution

Find the derivative of y with respect to x

To find dy/dt at any point, first find the derivative of y with respect to x (dy/dx) using the power rule and the constant rule. So the derivative of \(y = 2x^2 - 6x\) is \(dy/dx = 4x - 6\).

Find dy/dt when x = 3 and dx/dt = 2

Plug x = 3 into dy/dx to find the derivative of y at x = 3. This gives \(dy/dx = 4*3 - 6 = 6\). Then, use the chain rule to find dy/dt. According to the chain rule, \(dy/dt = dy/dx * dx/dt\) so \(dy/dt = 6*2 = 12\).

Rearrange the chain rule to find dx/dt

To find dx/dt, rearrange the chain rule formula to \(dx/dt = dy/dt / dy/dx\).

Find dx/dt when x = 1 and dy/dt = 5

Substitute x = 1 into dy/dx to find dy/dx when x = 1 giving \(dy/dx = 4*1 - 6 = -2\). Then, substitute dy/dt = 5 and dy/dx = -2 into the rearranged chain rule formula to get \(dx/dt = 5 / -2 = -2.5\)

Key concepts

Rate of Change

The rate of change refers to how a quantity changes over time. In the context of calculus, it often denotes how one variable changes with respect to another. For instance, if we think about a car moving along a highway, the speed of the car is the rate at which distance changes over time, also known as velocity.

Similarly, when dealing with functions in mathematics, the rate of change tells us how the output of a function changes as its input changes. This is crucial for understanding behaviors like speed and acceleration in physics, or growth rates in biology.

In the exercise problem, we explore the relationship between the variables through their derivatives: \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). These derivatives provide us with the rate of change of \( y \) and \( x \) with respect to time \( t \). Calculating these rates is essential to understanding how one variable impacts another as time progresses.

Chain Rule

The Chain Rule is a fundamental tool in calculus. It allows us to differentiate composite functions, essentially enabling us to find the derivative of a function within another function. It’s like peeling an onion, where each layer depends on the layer before it.

Mathematically, if you have two functions \( f \) and \( g \) such that \( y = f(g(x)) \), the chain rule states that the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). In simple terms, you're differentiating the outer function and multiplying it by the derivative of the inner function.

• In the exercise, the chain rule connects the rates of change of \( y \), \( x \), and \( t \): \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \).
• This tells us how the changes in \( x \) directly affect \( y \) over time through their dependency on each other.

Mastering the chain rule is vital for tackling complex derivative problems where variables are interdependent.

Derivative Calculation

The calculation of derivatives is a core aspect of calculus. A derivative essentially measures how a change in one quantity can bring about a change in another quantity. It's particularly useful when we want to know the slope or rate at which values are changing at any given point.

In the provided exercise, we calculate the derivative \( \frac{dy}{dx} \) using basic rules of differentiation: the power rule and the constant rule. The power rule tells us how to differentiate functions of the form \( x^n \), and is stated as \( \frac{d}{dx}x^n = nx^{n-1} \).

• For the given function \( y = 2x^2 - 6x \), applying this rule gives \( \frac{dy}{dx} = 4x - 6 \).
• This derivative indicates how \( y \) changes with respect to \( x \).

Once \( \frac{dy}{dx} \) is obtained, we can use it to find specific rates of change when combined with given information like \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). This interlinked scenario highlights the importance of derivative calculations in understanding variable relations in calculus.