Find an equation of the line tangent to the curve defined by " + 5xy +y = 68 point (2, 2) at the

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
icon
Related questions
icon
Concept explainers
Question

Show work please!

### Problem Statement: Tangent Line to a Curve

**Objective:**
Find an equation of the line tangent to the curve defined by the equation \( x^5 + 5xy + y^4 = 68 \) at the point \( (2, 2) \).

**Step-by-Step Solution:**

1. **Define the Function:**
   The given curve is \( x^5 + 5xy + y^4 = 68 \).

2. **Differentiation:**
   To find the tangent line, we need to determine the derivative of the function implicitly given the curve equation.

3. **Evaluate at Point:**
   We evaluate the slope at the given point \( (2, 2) \).

4. **Form the Tangent Line Equation:**
   Using the slope and the point \( (2, 2) \), we then write the equation of the tangent line in the form \( y = mx + b \).

**Interactive Calculation:**
There's a blank field indicating where users are expected to input the \( y \)-value obtained after working through the differentiation and evaluation steps.

### Detailed Explanation:

To solve this problem and find the equation of the tangent line, follow these steps:

1. **Implicit Differentiation:**
   - Differentiate both sides of the equation \( x^5 + 5xy + y^4 = 68 \) with respect to \( x \):

     \[
     \frac{d}{dx}(x^5) + \frac{d}{dx}(5xy) + \frac{d}{dx}(y^4) = \frac{d}{dx}(68)
     \]

   - Applying the product rule to \( 5xy \) and chain rule to \( y^4 \):

     \[
     5x^4 + 5 \left( x \frac{dy}{dx} + y \right) + 4y^3 \frac{dy}{dx} = 0
     \]

   - Group the terms involving \( \frac{dy}{dx} \):

     \[
     5x^4 + 5y + 5x \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = 0
     \]

     \[
     5x^4 + 5y = - \left( 5x + 4y
Transcribed Image Text:### Problem Statement: Tangent Line to a Curve **Objective:** Find an equation of the line tangent to the curve defined by the equation \( x^5 + 5xy + y^4 = 68 \) at the point \( (2, 2) \). **Step-by-Step Solution:** 1. **Define the Function:** The given curve is \( x^5 + 5xy + y^4 = 68 \). 2. **Differentiation:** To find the tangent line, we need to determine the derivative of the function implicitly given the curve equation. 3. **Evaluate at Point:** We evaluate the slope at the given point \( (2, 2) \). 4. **Form the Tangent Line Equation:** Using the slope and the point \( (2, 2) \), we then write the equation of the tangent line in the form \( y = mx + b \). **Interactive Calculation:** There's a blank field indicating where users are expected to input the \( y \)-value obtained after working through the differentiation and evaluation steps. ### Detailed Explanation: To solve this problem and find the equation of the tangent line, follow these steps: 1. **Implicit Differentiation:** - Differentiate both sides of the equation \( x^5 + 5xy + y^4 = 68 \) with respect to \( x \): \[ \frac{d}{dx}(x^5) + \frac{d}{dx}(5xy) + \frac{d}{dx}(y^4) = \frac{d}{dx}(68) \] - Applying the product rule to \( 5xy \) and chain rule to \( y^4 \): \[ 5x^4 + 5 \left( x \frac{dy}{dx} + y \right) + 4y^3 \frac{dy}{dx} = 0 \] - Group the terms involving \( \frac{dy}{dx} \): \[ 5x^4 + 5y + 5x \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = 0 \] \[ 5x^4 + 5y = - \left( 5x + 4y
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 4 images

Blurred answer
Knowledge Booster
Points, Lines and Planes
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning