Determine an equation of the line tangent to the curve at the given point.

x cubed plus y cubed equals 22 xyx3+y3=22xy​;

left parenthesis 11 comma 11 right parenthesis(11,11)

We are given the implicit equation \(x^3 + y^3 = 22xy\), and we need to find the equation of the line tangent to the curve at the point \((11, 11)\).

### Step 1: Differentiate the given equation implicitly
We will differentiate both sides of the equation with respect to \(x\) using implicit differentiation.

\[
\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(22xy)
\]

First, differentiate the left-hand side:

\[
\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = 3x^2 + 3y^2 \frac{dy}{dx}
\]

Now differentiate the right-hand side:

\[
\frac{d}{dx}(22xy) = 22\left( x \frac{dy}{dx} + y \right)
\]

So, the implicit differentiation gives us the following equation:

\[
3x^2 + 3y^2 \frac{dy}{dx} = 22\left( x \frac{dy}{dx} + y \right)
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
We want to isolate \(\frac{dy}{dx}\). First, expand both sides:

\[
3x^2 + 3y^2 \frac{dy}{dx} = 22x \frac{dy}{dx} + 22y
\]

Now, move all the terms involving \(\frac{dy}{dx}\) to one side:

\[
3y^2 \frac{dy}{dx} - 22x \frac{dy}{dx} = 22y - 3x^2
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx}(3y^2 - 22x) = 22y - 3x^2
\]

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

\[
\frac{dy}{dx} = \frac{22y - 3x^2}{3y^2 - 22x}
\]

### Step 3: Substitute the point \((11, 11)\)
Now, we substitute \(x = 11\) and \(y = 11\) into the derivative formula:

\[
\frac{dy}{dx} = \frac{22(11) - 3(11)^2}{3(11)^2 - 22(11)}
\]

Simplify the numerator and the denominator:

\[
\frac{dy}{dx} = \frac{242 - 363}{363 - 242} = \frac{-121}{121} = -1
\]

Thus, the slope of the tangent line at the point \((11, 11)\) is \(-1\).

### Step 4: Find the equation of the tangent line
The equation of the tangent line is given by the point-slope form of a line:

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

Substitute \(m = -1\), \(x_1 = 11\), and \(y_1 = 11\):

\[
y - 11 = -1(x - 11)
\]

Simplify the equation:

\[
y - 11 = -x + 11
\]

\[
y = -x + 22
\]

### Final Answer:
The equation of the tangent line at the point \((11, 11)\) is:

\[
y = -x + 22
\]

Would you like a detailed explanation of any step?

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Here are 5 questions to extend your understanding:

1. What is implicit differentiation and when do we use it?
2. How do you apply the point-slope form to find the equation of a tangent line?
3. Why is it important to isolate \(\frac{dy}{dx}\) when using implicit differentiation?
4. Can the tangent line at other points on the curve be steeper or less steep than at \((11, 11)\)?
5. How would you find where the slope of the tangent line is horizontal?

**Tip:** When performing implicit differentiation, remember to apply the product rule carefully when dealing with terms involving \(x\) and \(y\).

Determine an equation of the line tangent to the curve at the given point. x^3 + y^3 = 22xy; (11,11)

Find the equation of the tangent line to the curve x^3 + y^3 = 22xy at the point (11, 11) using implicit differentiation. This detailed solution explains how to apply implicit differentiation and the point-slope form of a line. Suitable for students in Grades 11-12.

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