Determine an equation of the line tangent to the curve at the given point.
32 left parenthesis x squared plus y squared right parenthesis squared equals 400 xy squared32x2+y22=400xy2​; left parenthesis 2 comma 4 right parenthesis(2,4)

We are tasked with finding the equation of the tangent line to the curve at the point \( (2, 4) \). The given equation is:

\[
32(x^2 + y^2)^2 = 400xy^2
\]

### Step 1: Implicit Differentiation
To find the slope of the tangent line, we need to differentiate both sides of the equation implicitly with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., \( y = y(x) \)).

#### Differentiate both sides:
\[
\frac{d}{dx}\left[ 32(x^2 + y^2)^2 \right] = \frac{d}{dx} \left[ 400xy^2 \right]
\]

1. Differentiate the left-hand side:

Using the chain rule, the derivative of \( 32(x^2 + y^2)^2 \) is:

\[
32 \cdot 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2)
\]

Now, differentiate \( x^2 + y^2 \):

\[
\frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx}
\]

So the derivative of the left-hand side becomes:

\[
64(x^2 + y^2)(x + y \frac{dy}{dx})
\]

2. Differentiate the right-hand side:

Using the product rule, the derivative of \( 400xy^2 \) is:

\[
\frac{d}{dx}(400xy^2) = 400 \left[ \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) \right]
\]

The derivative of \( x \) is 1, and the derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \). So, the derivative of the right-hand side becomes:

\[
400(y^2 + 2xy \frac{dy}{dx})
\]

### Step 2: Solve for \( \frac{dy}{dx} \)

Now we have the equation:

\[
64(x^2 + y^2)(x + y \frac{dy}{dx}) = 400(y^2 + 2xy \frac{dy}{dx})
\]

Substitute \( x = 2 \) and \( y = 4 \) into this equation.

1. Compute \( x^2 + y^2 \):

\[
x^2 + y^2 = 2^2 + 4^2 = 4 + 16 = 20
\]

2. Substitute \( x = 2 \), \( y = 4 \):

\[
64(20)(2 + 4 \frac{dy}{dx}) = 400(16 + 16 \frac{dy}{dx})
\]

Simplify both sides:

\[
1280(2 + 4 \frac{dy}{dx}) = 400(16 + 16 \frac{dy}{dx})
\]

Expand both sides:

\[
2560 + 5120 \frac{dy}{dx} = 6400 + 6400 \frac{dy}{dx}
\]

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

Rearrange the equation:

\[
5120 \frac{dy}{dx} - 6400 \frac{dy}{dx} = 6400 - 2560
\]

\[
-1280 \frac{dy}{dx} = 3840
\]

\[
\frac{dy}{dx} = \frac{3840}{-1280} = -3
\]

Thus, the slope of the tangent line at the point \( (2, 4) \) is \( -3 \).

### Step 4: Equation of the Tangent Line

We now use the point-slope form of the equation of a line:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = -3 \), \( x_1 = 2 \), and \( y_1 = 4 \):

\[
y - 4 = -3(x - 2)
\]

Simplify:

\[
y - 4 = -3x + 6
\]

\[
y = -3x + 10
\]

### Final Answer:
The equation of the tangent line is:

\[
y = -3x + 10
\]

Would you like more details, or do you have any questions about the solution?

---

Here are 5 related questions for further exploration:

1. How do you perform implicit differentiation with more complex functions?
2. What are the steps to apply the chain rule when differentiating implicitly?
3. How does the product rule apply to functions of two variables?
4. What is the geometric meaning of a tangent line to a curve at a given point?
5. How can implicit differentiation be extended to higher-order derivatives?

**Tip**: Always verify your derivative calculations step by step to avoid errors when working with implicit differentiation.

Determine an equation of the line tangent to the curve at the given point (2, 4) for the equation 32(x^2 + y^2)^2 = 400xy^2.

Discover how to find the equation of the tangent line to the curve 32(x^2 + y^2)^2 = 400xy^2 at the point (2, 4) using implicit differentiation. This detailed solution involves using the chain rule and product rule to calculate the slope and apply the point-slope formula. Perfect for Grade 12 or college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation