Problem 18 - Calculus Single Variable

Question

Variables \(x\) and \(y\) are functions of a parameter \(t\) and are related by a given equation. A point \(P_{0}\) on the graph of that equation is also given, as is one of the following two values: $$ v_{0}=\left.\frac{d x}{d t}\right|_{P_{0}} \quad \text { or } \quad s_{0}=\left.\frac{d y}{d t}\right|_{P_{0}} $$ Find the other value. Also, referring to the Insight that follows Example \(2,\) find the tangent line to the graph of the given equation at \(P_{0}\). $$ 2 x^{3}+y^{3}=8, \quad P_{0}=(2,-2), \quad v_{0}=-1 $$

Step-by-Step Solution

Answer

The unknown value is \( \frac{dy}{dt} = 2 \) and the tangent line is \( y = -2x + 2 \).

Step 1: Differentiate the given equation implicitly with respect to time

The given equation is \( 2x^3 + y^3 = 8 \). Differentiate both sides with respect to the parameter \( t \). Using implicit differentiation, we get \( 6x^2 \frac{dx}{dt} + 3y^2 \frac{dy}{dt} = 0 \).

Step 2: Substitute known values

We know the point \( P_0 = (2, -2) \) and \( v_0 = \left. \frac{dx}{dt} \right|_{P_0} = -1 \). Substitute these values into the differentiated equation to find \( s_0 = \left. \frac{dy}{dt} \right|_{P_0} \).

Step 3: Solve for \( \frac{dy}{dt} \)

Substitute \( x = 2 \), \( y = -2 \), and \( \frac{dx}{dt} = -1 \) into the differentiated equation: \[ 6(2)^2(-1) + 3(-2)^2 \frac{dy}{dt} = 0 \]Simplify to \( -24 + 12 \frac{dy}{dt} = 0 \). Solve for \( \frac{dy}{dt} \):\[ 12 \frac{dy}{dt} = 24 \] \[ \frac{dy}{dt} = 2 \].

Step 4: Find the derivative dy/dx for the tangent line

To find the slope of the tangent line to the graph at \( P_0 \), use the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute \( \frac{dy}{dt} = 2 \) and \( \frac{dx}{dt} = -1 \):\[ \frac{dy}{dx} = \frac{2}{-1} = -2 \].

Step 5: Write the equation of the tangent line

Use the point-slope form of a line equation \( y - y_1 = m(x - x_1) \), where \( m = -2 \) (the slope found), \( x_1 = 2 \), and \( y_1 = -2 \):\[ y + 2 = -2(x - 2) \]Simplify to get the tangent line equation:\[ y = -2x + 2 \].

Key Concepts

Tangent Line

When we discuss finding the tangent line to a curve at a specific point, we are essentially talking about finding a straight line that just touches the curve at that point. This line is special because it represents the slope of the curve at that precise moment and location. For example, at the point \( P_0 = (2, -2) \) on the graph given by the equation \( 2x^3 + y^3 = 8 \), the tangent line would have the same slope as the curve at \( P_0 \). The slope of the tangent is determined by the rate of change of \( y \) with respect to \( x \) at \( P_0 \), which we call \( \frac{dy}{dx} \).
To find the equation of the tangent line, we first calculate the slope using implicit differentiation, as we'll explain in more detail later.
Once we have the slope, we use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the tangent line, and \( x_1 \) and \( y_1 \) are the x and y coordinates of the point of tangency. In this case, the slope \( m \) was determined to be \(-2\), and substituting the known point \( (2, -2) \) into the equation gives us the line \( y = -2x + 2 \), which is the tangent line to the curve at \( P_0 \).

Parametric Equations

Parametric equations are a powerful way to describe geometric shapes and curves, such as the path of an object moving in two dimensions. These equations use a parameter, often \( t \), to express the x and y coordinates of points on the curve as functions of \( t \). This approach is highly useful when dealing with motion, as it naturally accommodates the idea of change over time.
In the given exercise, the relationship between \( x \) and \( y \) is tied through the parameter \( t \). The parametric form allows us to track how both \( x \) and \( y \) change as \( t \) changes, giving us a comprehensive view of the motion and behavior of the point on the graph.
The task in the exercise was to find the rate of change of \( y \) with respect to \( t \) at a specific point. By differentiating the equation \( 2x^3 + y^3 = 8 \) with respect to \( t \), we piece together how each individual variable changes over time. This process enables us to derive important information about the derivative \( \frac{dy}{dt} \), helping us understand the curve's behavior around the point \( P_0 \).

Derivative Calculation

Calculating derivatives is a core aspect of calculus and allows us to determine how one quantity changes in relation to another. This is crucial when analyzing motion and dynamic systems.
In implicit differentiation, we find the derivative of an equation where the dependent variable \( y \) is not isolated. Given the equation \( 2x^3 + y^3 = 8 \), we differentiate both sides with respect to the parameter \( t \), treating \( x \) and \( y \) as functions of \( t \).
During differentiation, we apply the chain rule, resulting in equations involving the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). After substituting known values from the specific point \( P_0 \), we solve for unknown rates.
For instance, in our exercise, we knew \( \frac{dx}{dt} = -1 \) and solved for \( \frac{dy}{dt} \) yielding \( 2 \). To find the slope of the tangent line, we use \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This gave us a slope of \(-2\), demonstrating how implicitly differentiated equations are used to find derivatives and slopes in parametric functions.