find the equation of the line tangent to 2xy=piesin y at the point (0,2pie)
To find the equation of the line tangent to the curve 2xy = πsin(y) at the given point (0, 2π), you can follow these steps:
Step 1: Differentiate the equation with respect to x using implicit differentiation.
Starting with the equation 2xy = πsin(y), differentiate both sides of the equation with respect to x. Remember that y is a function of x.
The derivative of 2xy with respect to x is 2y + 2xy', where y' denotes dy/dx (the derivative of y with respect to x).
The derivative of πsin(y) with respect to x is πcos(y)*y'.
Therefore, the differentiated equation becomes:
2y + 2xy' = πcos(y)*y'
Step 2: Substitute the given point into the equation.
We are given the point (0, 2π). Substitute x = 0 and y = 2π into the differentiated equation to find the value of y' (the slope of the tangent line).
2(2π) + 2(0)y' = πcos(2π)*y'
4π = πcos(2π)*y'
4π = π*(-1)*y'
4π = -πy'
y' = -4
Step 3: Write the equation of the tangent line in point-slope form.
Using the point-slope form of a linear equation (y-y1 = m(x-x1)), where m represents the slope and (x1, y1) represents the given point, we can substitute the values to find the equation of the tangent line:
y - 2π = -4(x - 0)
y - 2π = -4x
Therefore, the equation of the tangent line is y = -4x + 2π.