Find the slope of the line tangent to the curve 3x2y+cosy=3√+π2 at the point (−1,π6) .

Answer 1

To find the slope of the line tangent to the curve, we need to find the derivative of the curve at the point (-1, π/6).

First, we rearrange the equation of the curve to isolate y:
3x^2y + cos(y) = 3√x + π/2
3x^2y = -cos(y) + 3√x + π/2
y = (-cos(y) + 3√x + π/2) / (3x^2)

Now, we can find the derivative of y with respect to x using implicit differentiation.

Taking the derivative of both sides of the equation with respect to x:

d/dx [y] = d/dx [(-cos(y) + 3√x + π/2) / (3x^2)]

Using the quotient rule on the right side:

dy/dx = [(-sin(y)dy/dx) + (3/(2√x))]/(3x^2) - (2(-cos(y) + 3√x + π/2))/(3x^3)

At the point (-1, π/6), we substitute x=-1 and y=π/6 into the derivative:

dy/dx = [(-sin(π/6)dy/dx) + (3/(2√(-1)))]/((-1)^2) - (2(-cos(π/6) + 3√(-1) + π/2))/((-1)^3)
dy/dx = [(-1/2)dy/dx + (3/(2i))] - (2(-√3/2 + 3i + π/2))/(-1)
dy/dx = (-1/2)dy/dx + (3/(2i)) + √3/2 - 3i - π/2

Now, we rearrange the equation to solve for dy/dx (the slope):

(1/2)dy/dx = (3/(2i)) + √3/2 - 3i - π/2
dy/dx = 2[(3/(2i)) + √3/2 - 3i - π/2]
dy/dx = (3i - 3√3 + 4i - 2π)/i
dy/dx = (3i + 4i - 3√3 - 2π)/i
dy/dx = (7i - 3√3 - 2π)/i

The slope of the line tangent to the curve at the point (-1, π/6) is given by the imaginary part of the expression (7i - 3√3 - 2π)/i:

slope = -3√3 - 2π