I'm not sure how to go about this problem:
Find the equation of the line tangent to the curve x^2 + 2xy + y^3 = 4 at y=1 in the 1st quadrant.
first find the point where y = 1
x^2 + 2x(1) + 1 = 4
x^2 + 2x - 3 = 0
(x+3)(x-1) = 0
x = -3 or x = 1
but in quad I, the point is (1,1)
use implicit differentiation:
2x + 2x(dy/dx) + 2y + 3y^2 (dy/dx) = 0
dy/dx (2x + 3y^2) = -2x - 2y
dy/dx = (-2x - 2y)/(2x + 3y^2)
= (-2-2)/(2+3) = -4/5
equation of tangent:
y - 1 = (-4/5)(x-1)
5y - 5 = -4x + 4
4x + 5y = 9
check:
http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+2xy+%2B+y%5E3+%3D+4,+4x+%2B+5y+%3D+9
see what happens at (1,1)
Thank you so much :)
To find the equation of the line tangent to the curve at a given point, you need to find the slope of the tangent line at that point and then use the point-slope form of a line to determine the equation.
To find the slope of the tangent line at a point on the curve, you can use implicit differentiation. In this case, we need to find the derivative of the equation x^2 + 2xy + y^3 = 4 with respect to x.
Let's start by differentiating both sides of the equation with respect to x:
d/dx(x^2 + 2xy + y^3) = d/dx(4)
Differentiating each term separately, we get:
2x + 2y * dy/dx + 3y^2 * dy/dx = 0
Now, we need to find dy/dx, which represents the slope of the tangent line. To do this, we can rearrange the equation to solve for dy/dx:
2y * dy/dx + 3y^2 * dy/dx = -2x
(2y + 3y^2) * dy/dx = -2x
dy/dx = -2x / (2y + 3y^2)
Now we have the slope of the tangent line at any point (x, y) on the curve given by -2x / (2y + 3y^2). We can substitute y = 1 (as given in the problem) to find the slope at y = 1 in the 1st quadrant:
m = -2x / (2(1) + 3(1)^2)
m = -2x / (2 + 3)
m = -2x / 5
Now that we have the slope, we can use the point-slope form of a line to find the equation. The point-slope form is given by:
y - y1 = m(x - x1)
Substituting the given point (x1, y1) = (1, 1) and the slope m = -2x/5, we get:
y - 1 = (-2x/5)(x - 1)
Simplifying the equation, we have:
y - 1 = (-2/5)x + 2/5
Finally, rearranging the equation to slope-intercept form (y = mx + b), we get:
y = (-2/5)x + 7/5
Therefore, the equation of the line tangent to the curve x^2 + 2xy + y^3 = 4 at y = 1 in the 1st quadrant is y = (-2/5)x + 7/5.