how would i go about finding the tangent line to the curve of sqrt(1+4x^2y^2)-3xy=-18 with the points (3,6)?
find slope at (3,6)
show line with that slope though (3,6)
√(1+4x^2y^2)-3xy = -18
1/2√(1+4x^2y^2) * (8xy^2 + 8x^2y y') - 2y - 2xy' = 0
8xy^2/2√(1+4x^2y^2) - 2y = y'(8x^2y/2√(1+4x^2y^2) - 2x) = 0
y' = (2y - 8xy^2/2√(1+4x^2y^2) )/(8x^2y/2√(1+4x^2y^2) - 2x)
plug in (3,6) for (x,y) and evaluate for slope.
find line.
To find the tangent line to a curve at a given point, you need to find the derivative of the curve and then substitute the coordinates of the point into the derivative equation to find the slope of the tangent line. Finally, you can use the point-slope form of a line to write the equation of the tangent line.
Here are the steps to find the tangent line to the curve given by the equation √(1+4x^2y^2) - 3xy = -18 at the point (3, 6):
Step 1: Differentiate the equation
Take the derivative of both sides of the equation with respect to "x." Treat "y" as a function of "x" while differentiating.
The derivative of √(1+4x^2y^2) with respect to "x" can be found using the chain rule. The derivative of √(1+4x^2y^2) is (1/2) * (1+4x^2y^2)^(-1/2) * 8x^2y * y'.
Differentiating -3xy with respect to "x" gives -3y - 3xy'.
Putting it all together, we have:
(1/2) * (1+4x^2y^2)^(-1/2) * 8x^2y * y' - 3y - 3xy' = 0
Step 2: Substitute the coordinates of the point
Substitute x = 3 and y = 6 into the equation obtained from Step 1. Solve for y' (the derivative of y with respect to x).
(1/2) * (1+4(3)^2(6)^2)^(-1/2) * 8(3)^2(6) * y' - 3(6) - 3(3)(y') = 0
Simplifying the expression leads to:
(1/2) * (1+432)^(-1/2) * 432y' - 18 - 9y' = 0
Step 3: Solve for y'
Rearrange the equation to solve for y':
(1/2) * (433)^(-1/2) * 432y' - 18 - 9y' = 0
(1/2) * (433)^(-1/2) * 432y' - 9y' = 18
[(1/2) * (433)^(-1/2) * 432 - 9]y' = 18
[(1/2) * (433)^(-1/2) * 432 - 9]y' = 18
Divide both sides by [(1/2) * (433)^(-1/2) * 432 - 9] to solve for y'.
y' = 18 / [(1/2) * (433)^(-1/2) * 432 - 9]
Step 4: Find the slope of the tangent line
Evaluate the expression obtained in Step 3 to find the slope of the tangent line.
Substitute the values of x = 3, y = 6, and the expression for y' into the equation.
slope = y' = 18 / [(1/2) * (433)^(-1/2) * 432 - 9]
Step 5: Write the equation of the tangent line
Finally, use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) are the coordinates of the given point and m is the slope obtained in Step 4.
Substitute the values of x = 3, y = 6 and the slope into the equation.
y - 6 = slope * (x - 3)
This is the equation of the tangent line to the curve at the point (3, 6).