Find the equation of the tangent line to the curve x^3 + xy^2 =5 at the point (1,2).
To find the equation of the tangent line to the curve at the point (1,2), we can use the concept of derivatives.
Step 1: Take the derivative of the equation with respect to x. This will give us the slope of the tangent line at any given point on the curve.
Differentiating the equation x^3 + xy^2 = 5 with respect to x gives:
3x^2 + y^2 + 2xy(dy/dx) = 0
Step 2: Plug in the coordinates of the given point (1,2) into the derivative equation to find the value of dy/dx at that point.
Substitute x = 1 and y = 2 into the equation:
3(1)^2 + (2)^2 + 2(1)(2)(dy/dx) = 0
3 + 4 + 4(dy/dx) = 0
7 + 4(dy/dx) = 0
4(dy/dx) = -7
dy/dx = -7/4
Step 3: Use the slope of the tangent line (dy/dx) and the coordinates of the point (1,2) to write the equation of the tangent line in point-slope form.
The point-slope form of a line is given by: y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope.
Using the point (1,2) and the slope (-7/4), we have:
y - 2 = (-7/4)(x - 1)
Simplifying the equation:
y - 2 = (-7/4)x + (7/4)
y = (-7/4)x + (7/4) + 2
y = (-7/4)x + (15/4)
Therefore, the equation of the tangent line to the curve x^3 + xy^2 = 5 at the point (1,2) is y = (-7/4)x + (15/4).