Find The Slope Of The Tangent Line To The Implicit Curve $x^{2/3} + Y^{2/3} - Xy = 0$ At The Point $(1, 3.148)$.SHOW ALL STEPS And Round Your Answer To 3 Decimal Places.

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Introduction

Implicit differentiation is a powerful technique used to find the derivative of an implicitly defined function. In this article, we will use implicit differentiation to find the slope of the tangent line to the implicit curve $x^{2/3} + y^{2/3} - xy = 0$ at the point $(1, 3.148)$. This technique is essential in calculus and has numerous applications in various fields, including physics, engineering, and economics.

Implicit Differentiation

Implicit differentiation is a method of differentiating both sides of an equation with respect to a variable, usually x, while treating the other variable, usually y, as a function of x. This technique is used when the equation is not easily solvable for y.

To differentiate an implicit equation, we will use the following rules:

  • If y is a function of x, then the derivative of y with respect to x is denoted as $\frac{dy}{dx}$.
  • If x is a function of y, then the derivative of x with respect to y is denoted as $\frac{dx}{dy}$.
  • The derivative of a product of two functions is the product of the derivatives of the individual functions, i.e., $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.
  • The derivative of a quotient of two functions is the quotient of the derivatives of the individual functions, i.e., $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Differentiating the Implicit Curve

We are given the implicit curve $x^{2/3} + y^{2/3} - xy = 0$. To find the slope of the tangent line at the point $(1, 3.148)$, we will use implicit differentiation.

First, we will differentiate both sides of the equation with respect to x:

ddx(x2/3+y2/3−xy)=ddx(0)\frac{d}{dx}\left(x^{2/3} + y^{2/3} - xy\right) = \frac{d}{dx}(0)

Using the power rule for differentiation, we get:

23x−1/3+23y−1/3dydx−y−xdydx=0\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} - y - x\frac{dy}{dx} = 0

Solving for $\frac{dy}{dx}$

Now, we will solve for $\frac{dy}{dx}$:

23y−1/3dydx−xdydx=y−23x−1/3\frac{2}{3}y^{-1/3}\frac{dy}{dx} - x\frac{dy}{dx} = y - \frac{2}{3}x^{-1/3}

Factoring out $\frac{dy}{dx}$, we get:

(23y−1/3−x)dydx=y−23x−1/3\left(\frac{2}{3}y^{-1/3} - x\right)\frac{dy}{dx} = y - \frac{2}{3}x^{-1/3}

Dividing both sides by $\left(\frac{2}{3}y^{-1/3} - x\right)$, we get:

dydx=y−23x−1/323y−1/3−x\frac{dy}{dx} = \frac{y - \frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3} - x}

Evaluating the Derivative at the Given Point

Now, we will evaluate the derivative at the point $(1, 3.148)$:

dydx=3.148−23(1)−1/323(3.148)−1/3−1\frac{dy}{dx} = \frac{3.148 - \frac{2}{3}(1)^{-1/3}}{\frac{2}{3}(3.148)^{-1/3} - 1}

Simplifying the expression, we get:

dydx=3.148−223(3.148)−1/3−1\frac{dy}{dx} = \frac{3.148 - 2}{\frac{2}{3}(3.148)^{-1/3} - 1}

dydx=1.14823(3.148)−1/3−1\frac{dy}{dx} = \frac{1.148}{\frac{2}{3}(3.148)^{-1/3} - 1}

Calculating the Numerical Value

To calculate the numerical value of the derivative, we will use a calculator:

dydx=1.14823(3.148)−1/3−1≈1.1480.148≈7.783\frac{dy}{dx} = \frac{1.148}{\frac{2}{3}(3.148)^{-1/3} - 1} \approx \frac{1.148}{0.148} \approx 7.783

Rounding the answer to 3 decimal places, we get:

dydx≈7.783\frac{dy}{dx} \approx 7.783

Conclusion

In this article, we used implicit differentiation to find the slope of the tangent line to the implicit curve $x^{2/3} + y^{2/3} - xy = 0$ at the point $(1, 3.148)$. We differentiated both sides of the equation with respect to x, solved for $\frac{dy}{dx}$, and evaluated the derivative at the given point. The final answer is $\frac{dy}{dx} \approx 7.783$.

References

  • [1] Anton, H. (2010). Calculus: Early Transcendentals. John Wiley & Sons.
  • [2] Edwards, C. H. (2010). Calculus. Pearson Education.
  • [3] Larson, R. (2010). Calculus. Cengage Learning.

Note: The references provided are for general calculus textbooks and are not specific to the problem at hand. However, they provide a good starting point for further reading and understanding of the concepts used in this article.

Introduction

Implicit differentiation is a powerful technique used to find the derivative of an implicitly defined function. In this article, we will answer some frequently asked questions about implicit differentiation, including its applications, limitations, and common mistakes.

Q: What is implicit differentiation?

A: Implicit differentiation is a method of differentiating both sides of an equation with respect to a variable, usually x, while treating the other variable, usually y, as a function of x. This technique is used when the equation is not easily solvable for y.

Q: When is implicit differentiation used?

A: Implicit differentiation is used when the equation is not easily solvable for y, or when the equation is a function of multiple variables. It is commonly used in physics, engineering, and economics to model real-world problems.

Q: What are the steps involved in implicit differentiation?

A: The steps involved in implicit differentiation are:

  1. Differentiate both sides of the equation with respect to x.
  2. Use the power rule for differentiation to differentiate the terms.
  3. Solve for $\frac{dy}{dx}$.
  4. Evaluate the derivative at the given point.

Q: What are some common mistakes to avoid in implicit differentiation?

A: Some common mistakes to avoid in implicit differentiation include:

  • Forgetting to differentiate both sides of the equation.
  • Not using the power rule for differentiation.
  • Not solving for $\frac{dy}{dx}$ correctly.
  • Not evaluating the derivative at the given point.

Q: What are some applications of implicit differentiation?

A: Some applications of implicit differentiation include:

  • Finding the slope of a tangent line to a curve.
  • Finding the equation of a tangent line to a curve.
  • Modeling real-world problems in physics, engineering, and economics.

Q: What are some limitations of implicit differentiation?

A: Some limitations of implicit differentiation include:

  • It can be difficult to solve for $\frac{dy}{dx}$ in some cases.
  • It may not be possible to find an explicit expression for y.
  • It may not be possible to evaluate the derivative at a given point.

Q: How do I choose between implicit differentiation and explicit differentiation?

A: To choose between implicit differentiation and explicit differentiation, you should consider the following:

  • If the equation is easily solvable for y, use explicit differentiation.
  • If the equation is not easily solvable for y, use implicit differentiation.
  • If the equation is a function of multiple variables, use implicit differentiation.

Q: Can I use implicit differentiation to find the derivative of a function that is not implicitly defined?

A: No, implicit differentiation is only used to find the derivative of an implicitly defined function. If the function is not implicitly defined, you should use explicit differentiation.

Q: Can I use implicit differentiation to find the derivative of a function that is defined in terms of a parameter?

A: Yes, you can use implicit differentiation to find the derivative of a function that is defined in terms of a parameter. However, you should be careful to differentiate the parameter correctly.

Q: How do I evaluate the derivative at a given point using implicit differentiation?

A: To evaluate the derivative at a given point using implicit differentiation, you should:

  1. Substitute the given point into the equation.
  2. Differentiate both sides of the equation with respect to x.
  3. Solve for $\frac{dy}{dx}$.
  4. Evaluate the derivative at the given point.

Conclusion

Implicit differentiation is a powerful technique used to find the derivative of an implicitly defined function. It has numerous applications in physics, engineering, and economics, and is commonly used to model real-world problems. However, it can be difficult to solve for $\frac{dy}{dx}$ in some cases, and may not be possible to find an explicit expression for y. By following the steps involved in implicit differentiation and avoiding common mistakes, you can use this technique to find the derivative of an implicitly defined function.

References

  • [1] Anton, H. (2010). Calculus: Early Transcendentals. John Wiley & Sons.
  • [2] Edwards, C. H. (2010). Calculus. Pearson Education.
  • [3] Larson, R. (2010). Calculus. Cengage Learning.

Note: The references provided are for general calculus textbooks and are not specific to the problem at hand. However, they provide a good starting point for further reading and understanding of the concepts used in this article.