Problem 20 - Calculus Single Variable

Question

Use implicit differentiation to find the tangent line to the given curve at the given point \(P_{0}\). \(x+y / x-\exp \left(y^{2}\right)=2 \quad P_{0}=(3,0)\)

Step-by-Step Solution

Answer

The tangent line equation is \(y = -3x + 9\).

Step 1: Implicit Differentiation Setup

Start by writing the given equation: \(x + \frac{y}{x} - e^{y^2} = 2\). You need to differentiate both sides with respect to \(x\) using implicit differentiation. Keep in mind that \(y\) is a function of \(x\), so whenever you differentiate a term with \(y\), you must multiply by \(\frac{dy}{dx}\).

Step 2: Differentiate Each Term

Differentiate each term:- The derivative of \(x\) is 1.- The derivative of \(\frac{y}{x}\) using the quotient rule is \(\frac{x \cdot \frac{dy}{dx} - y}{x^2}\).- The derivative of \(e^{y^2}\) using the chain rule is \(e^{y^2} \cdot 2y \cdot \frac{dy}{dx}\).- The derivative of the constant 2 is 0. Set the differentiated equation: \[1 + \frac{x \cdot \frac{dy}{dx} - y}{x^2} - e^{y^2} \cdot 2y \cdot \frac{dy}{dx} = 0\].

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

Rearrange the equation to isolate \(\frac{dy}{dx}\):\[1 = -\frac{x \cdot \frac{dy}{dx} - y}{x^2} + e^{y^2} \cdot 2y \cdot \frac{dy}{dx}\].\[x^2 = -x \cdot \frac{dy}{dx} + y + x^2 e^{y^2} \cdot 2y \cdot \frac{dy}{dx}\].Collect terms with \(\frac{dy}{dx}\): \[(x + x^2 e^{y^2} \cdot 2y)\frac{dy}{dx} = y - x^2\].Finally, solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{y - x^2}{x + x^2 e^{y^2} \cdot 2y}\].

Step 4: Evaluate at Given Point

Substitute \(x = 3\) and \(y = 0\) into the derivative to find the slope:\[\frac{dy}{dx} = \frac{0 - 9}{3 + 3^2 \cdot e^{0} \cdot 2 \cdot 0}\]. This simplifies to \[\frac{dy}{dx} = \frac{-9}{3}\].Hence, \(\frac{dy}{dx} = -3\).

Step 5: Form the Tangent Line Equation

Use the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step. With \((x_1, y_1) = (3, 0)\) and slope \(-3\):\[y - 0 = -3(x - 3)\].Simplify to \(y = -3x + 9\).

Key Concepts

Tangent Line

The concept of a tangent line is fundamental when dealing with curves and their slopes at specific points. A tangent line is a straight line that touches a curve at exactly one point without crossing it. This line represents the instantaneous rate of change of the curve at that particular point. It's like an arrow showing the direction the curve is heading at that point.
To find the equation of this line, we need to know both a point on the line and the slope of the line at that point. In our exercise, we find the tangent line using implicit differentiation. Once we have the slope, we can use the point-slope form of a line to write its equation. This results in a clear graphical representation of how the curve behaves near the point of tangency.

Quotient Rule

When differentiating functions involving division, the quotient rule is invaluable. It's a technique used to find the derivative of a function that is a ratio of two differentiable functions. If you have a function represented as \ \( \frac{u(x)}{v(x)} \ \), its derivative can be found using:

• \( (u/v)' = (v\cdot u' - u\cdot v')/v^2 \).

In our specific case, we differentiate the term \( \frac{y}{x} \). Here, \( u = y \) and \( v = x \), where \( u' = \frac{dy}{dx} \) and \( v' = 1 \). Applying the quotient rule helps us find \( \frac{x \cdot \frac{dy}{dx} - y}{x^2} \) for part of the differentiated equation. This rule allows us to accurately compute the slope of the curve at the required point.

Chain Rule

The chain rule is another essential differentiation technique, especially when dealing with composite functions. When you have a function composed within another function, the chain rule allows you to differentiate it. It's particularly helpful for complex functions like \( e^{y^2} \) in our given exercise.The general idea is: if you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). In our case, this becomes:

• \( \frac{d}{dx}[e^{y^2}] = e^{y^2} \cdot 2y \cdot \frac{dy}{dx} \).

We multiply by \( 2y \) because of the derivative of \( y^2 \). This illustrates how inner functions contribute to the derivative of the entire function, ensuring continuity and correctness in our calculations.

Derivative

The derivative is a cornerstone in calculus, representing the rate at which a function changes at a specific point. It's essentially the slope of the function at any given point. Differentiation, the process of finding a derivative, helps us determine how one quantity changes in relation to another.In our exercise, the need to differentiate implicitly arises because \( y \) is a function of \( x \) hidden within the given equation. By applying derivatives to each term, including constants, products, or quotients, we uncover the relationship between \( x \) and \( y \) through \( \frac{dy}{dx} \). This derivative gives us the slope of the tangent line at \( P_0 = (3,0) \), allowing us to understand how the curve bends and changes there.

Point-Slope Form

The point-slope form is a widely used method to write the equation of a line when you know a specific point on the line and the slope. It is expressed as:

• \( y - y_1 = m(x - x_1) \).

Here, \( m \) is the slope, and \( (x_1, y_1) \) is the point on the line. This form is especially useful in calculus and analytical geometry because it provides a straightforward way to create linear equations once the slope and a point are known.For the tangent line in our problem, with the point \( (3,0) \) and slope \( -3 \), substitution into the point-slope form gives us the equation \( y = -3x + 9 \). This equation not only describes the tangent line but also serves as a tool to inspect or predict behavior near \( P_0 \). It's a perfect illustration of turning calculus computations into practical equations that describe real-world scenarios.