Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

4. Applications of Derivatives

Implicit Differentiation

Calculus

4. Applications of Derivatives

Implicit Differentiation

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2:14 minutes

Problem 3.8.50a

Textbook Question

45–50. Tangent lines Carry out the following steps. <IMAGE>
a. Verify that the given point lies on the curve.
(x²+y²)²=25/4 xy²; (1, 2)

Verified step by step guidance

First, substitute the given point (1, 2) into the equation of the curve \((x^2 + y^2)^2 = \frac{25}{4}xy^2\).

Calculate \(x^2 + y^2\) by substituting \(x = 1\) and \(y = 2\) into the expression, resulting in \(1^2 + 2^2 = 1 + 4 = 5\).

Substitute \(x = 1\) and \(y = 2\) into the right side of the equation \(\frac{25}{4}xy^2\), which becomes \(\frac{25}{4} \times 1 \times 2^2 = \frac{25}{4} \times 4 = 25\).

Now, substitute the calculated values into the original equation: \((5)^2 = 25\) and \(25 = 25\).

Since both sides of the equation are equal, the point (1, 2) lies on the curve.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In this case, we differentiate both sides of the equation (x² + y²)² = 25/4 xy² with respect to x, treating y as a function of x. This allows us to find the slope of the tangent line at a specific point on the curve.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point and has the same slope as the curve at that point. The slope of the tangent line can be found using the derivative of the function at that point. The equation of the tangent line can then be expressed in point-slope form, using the coordinates of the point and the calculated slope.

Verifying Points on Curves

To verify that a point lies on a curve defined by an equation, we substitute the coordinates of the point into the equation. If the left-hand side equals the right-hand side after substitution, the point is confirmed to be on the curve. In this case, substituting (1, 2) into the equation (x² + y²)² = 25/4 xy² will determine if the point lies on the curve.