Use implicit differentiation to find an equation of the tangent line to the curve at the given point x^2 + xy + y^2 = 3, (1,1)
2x + y + xy' + 2yy' = 0
You know x,y so just solve for y'.
differentiate x^2+xy+y^2
f'= 2x+y+xy'+2yy'
f' =2(1)+(1)+(1)y'+2(1)y'
f'= 3+3y'
-3=3y'
y'=-1
to find tangent line,
y-y1=m(x-x1)
y-1= -1(x-1)
y-1= -x+1
x+y= 1+1
x+y=2
answer: x+y=2
To find the equation of the tangent line to the curve at the point (1,1), we need to differentiate both sides of the given equation with respect to x using the chain rule.
1. Start with the given equation: x^2 + xy + y^2 = 3.
2. Differentiate both sides of the equation with respect to x:
d/dx(x^2) + d/dx(xy) + d/dx(y^2) = d/dx(3).
3. Applying the power rule for differentiation, we get:
2x + x(dy/dx) + y + 2y(dy/dx) = 0.
4. Rearrange the equation to solve for dy/dx, which represents the slope of the tangent line:
x(dy/dx) + 2y(dy/dx) = -2x - y.
5. Factor out dy/dx on the left-hand side:
(x + 2y)(dy/dx) = -2x - y.
6. Solve for dy/dx by dividing both sides by (x + 2y):
dy/dx = (-2x - y) / (x + 2y).
7. Substitute the coordinates of the given point (1,1) into the equation:
dy/dx = (-2(1) - 1) / (1 + 2(1)).
dy/dx = (-2 - 1) / (1 + 2).
dy/dx = -3 / 3 = -1.
8. Therefore, the slope of the tangent line at the point (1,1) is -1.
9. Finally, we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y₁ = m(x - x₁),
where (x₁, y₁) represents the point (1,1) and m represents the slope (-1).
Plugging in the values, we get:
y - 1 = -1(x - 1).
Simplifying, we get:
y - 1 = -x + 1.
Rearranging the equation, we get the equation of the tangent line in slope-intercept form:
y = -x + 2.
Therefore, the equation of the tangent line to the curve x^2 + xy + y^2 = 3 at the point (1,1) is y = -x + 2.
To find the equation of the tangent line to a curve at a given point using implicit differentiation, follow these steps:
Step 1: Start with the equation of the curve.
Given: x^2 + xy + y^2 = 3
Step 2: Differentiate both sides of the equation with respect to x. Treat y as a function of x and apply the product rule for differentiating xy.
Differentiate: d/dx (x^2 + xy + y^2) = d/dx(3)
Differentiate each term:
2x + x(dy/dx) + y + 2y(dy/dx) = 0
Step 3: Simplify the equation and collect the terms involving dy/dx on one side.
Rearrange the terms:
(x + 2y)(dy/dx) = -2x - y
Now, divide both sides by (x + 2y):
dy/dx = (-2x - y)/(x + 2y)
Step 4: Substitute the given point (1,1) into the equation to find the slope of the tangent line at that point.
Substitute x = 1 and y = 1 into the equation:
dy/dx = (-2(1) - 1)/(1 + 2(1))
dy/dx = (-3)/(3)
dy/dx = -1
Step 5: 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)
Substitute the slope (m = -1) and the point (x1 = 1, y1 = 1) into the equation:
y - 1 = -1(x - 1)
Simplify:
y - 1 = -x + 1
Rearrange to get the equation in slope-intercept form:
y = -x + 2
Therefore, the equation of the tangent line to the curve at the point (1,1) is y = -x + 2.