Find an equation of the line tangent to the curve defined by x3+2xy+y3=4 at the point (1,1).

The slope of the tangent line at any point is y'

3x^2 + 2y + 2xy' + 3y^2 y' = 0
y' = -(3x^2+2y)/(2x+3y^2)
At (1,1) y' = -1
So the line is
y-1 = -1(x-1)

See the graphs at
www.wolframalpha.com/input/?i=plot+x%5E3%2B2xy%2By%5E3%3D4%2C+y+%3D+2-x

Why did the line go to the therapist? Because it needed some slope support!

Now, to find the equation of the tangent line, we'll need to find the derivative of the given curve and evaluate it at the point (1,1). Let's get down to business!

Differentiating x^3 + 2xy + y^3 = 4 with respect to x gives us:

3x^2 + 2x(dy/dx) + 2y + 3y^2(dy/dx) = 0

Now, let's substitute x = 1 and y = 1 into the equation:

3(1)^2 + 2(1)(dy/dx) + 2(1) + 3(1)^2(dy/dx) = 0

Simplifying this equation, we get:

3 + 2(dy/dx) + 2 + 3(dy/dx) = 0

Combining like terms, we have:

5(dy/dx) = -5

Dividing both sides by 5, we get:

dy/dx = -1

So, the slope of the tangent line at the point (1,1) is -1.

To find the equation of the line, we use the point-slope form:

y - y₁ = m(x - x₁)

Plugging in the values, we get:

y - 1 = -1(x - 1)

Simplifying this, we have:

y - 1 = -x + 1
y = -x + 2

Therefore, the equation of the line tangent to the curve at the point (1,1) is y = -x + 2.

To find the equation of the line tangent to the curve at the point (1,1), we need to find the slope of the tangent line at that point.

First, let's find the derivative of the equation x^3 + 2xy + y^3 = 4 with respect to x.

Taking the derivative of both sides of the equation, we get:

3x^2 + 2y + 2xy' + 3y^2y' = 0

Since we are looking for the slope at the point (1,1), substitute x = 1 and y = 1 into the equation:

3(1)^2 + 2(1) + 2(1)(y') + 3(1)^2(y') = 0

Simplifying further:

3 + 2 + 2y' + 3y' = 0

5y' = -5

y' = -1

Therefore, the slope of the tangent line at the point (1,1) is -1.

Now, we have the point (1,1) and the slope -1. We can use the point-slope form of a line to find the equation of the tangent line.

The point-slope form of a line is:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is a point on the line.

Substituting the values, we have:

y - 1 = -1(x - 1)

Simplifying further:

y - 1 = -x + 1

y = -x + 2

Therefore, the equation of the line tangent to the curve defined by x^3 + 2xy + y^3 = 4 at the point (1,1) is y = -x + 2.

To find the equation of the line tangent to the curve at the point (1,1), we can use differentiation.

1. Start by differentiating both sides of the equation.
The derivative of x^3 with respect to x is 3x^2.
The derivative of 2xy with respect to x is 2y + 2x(dy/dx).
The derivative of y^3 with respect to x is 3y^2(dy/dx).

2. Using implicit differentiation, differentiate each term with respect to x.

d/dx(x^3) + d/dx(2xy) + d/dx(y^3) = d/dx(4)

3x^2 + 2y + 2x(dy/dx) + 3y^2(dy/dx) = 0

3. Plug in the point (1,1) into the equation.
At (1,1), x = 1, y = 1.

3(1)^2 + 2(1) + 2(1)(dy/dx) + 3(1)^2(dy/dx) = 0

3 + 2 + 2(dy/dx) + 3(dy/dx) = 0

5 + 5(dy/dx) = 0

4. Solve for dy/dx.
Rearrange the equation and isolate dy/dx.

5(dy/dx) = -5

dy/dx = -5/5 = -1

5. Now we have the slope of the tangent line, which is dy/dx = -1.

6. To find the y-intercept of the line, substitute the coordinates (1,1) into the equation y = mx + c.
Using the point (1,1) and slope -1:

1 = -1(1) + c

1 = -1 + c

c = 2

7. The equation of the tangent line is y = -x + 2.

So, the equation of the line tangent to the curve defined by x^3 + 2xy + y^3 = 4 at the point (1,1) is y = -x + 2.