Find the equations of the tangent lines to the following curves at the indicated points.
xy^2 = 1 at (1, −1)
xy^2 = 1
x(2y)dy/dx + y^2 (1) = 0
dy/dx = -y^2/(2xy)
at (1,-1) , dy/dx = -(-1)^2/(2(1)(-1))= 1/-2 = - 1/2
Now you have the slope and the given point. Find the
equation of the straight line, using your method of choice.
To find the equations of the tangent lines to the curve xy^2 = 1 at the point (1, -1), we can follow these steps:
Step 1: Differentiate the equation implicitly with respect to x.
Taking the derivative of both sides with respect to x will yield the slope of the tangent line at any given point (x, y) on the curve. The derivative of xy^2 = 1 can be found using the product rule of differentiation:
d(xy^2)/dx = d(1)/dx
Using the product rule, the left side can be written as:
y^2 * dx/dx + x * d(y^2)/dx = 0
Since dx/dx is 1, we can simplify the equation to:
y^2 + 2xy * dy/dx = 0
Step 2: Substitute the coordinates of the given point into the equation.
To find the specific equation of the tangent line at the point (1, -1), we need to substitute x = 1 and y = -1 into the equation derived in step 1:
(-1)^2 + 2(1)(-1) * dy/dx = 0
1 - 2 * dy/dx = 0
Step 3: Solve for dy/dx.
Rearranging the equation from step 2, we can solve for dy/dx:
2 * dy/dx = 1
dy/dx = 1/2
Step 4: Find the equation of the tangent line using the point-slope form.
Now that we have the slope, we can use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values into the equation:
y - (-1) = (1/2)(x - 1)
Simplifying:
y + 1 = (1/2)(x - 1)
y = (1/2)x - 1/2 - 1
y = (1/2)x - 3/2
Therefore, the equation of the tangent line to the curve xy^2 = 1 at the point (1, -1) is y = (1/2)x - 3/2.