if tangent to the curve y=2sinx +sin2x are drawn at p(x=60degree) then find the area of the quadrilateral formed by the tangent the normal at P and the coordinate axis
To find the area of the quadrilateral formed by the tangent, the normal at point P, and the coordinate axes, follow these steps:
1. Find the coordinates of point P.
- Given x = 60 degrees, substitute this value into the equation y = 2sin(x) + sin(2x) to find the corresponding y-coordinate.
- Calculate y = 2sin(60°) + sin(2 * 60°) = 2(√3/2) + √3 ≈ 0.732.
2. Find the slope of the tangent line at point P.
- Differentiate the equation y = 2sin(x) + sin(2x) with respect to x.
- dy/dx = 2cos(x) + 2cos(2x) (using the chain rule).
- Plug x = 60 degrees into the derivative, dy/dx = 2cos(60°) + 2cos(2 * 60°) = 1 + 2(1/2) = 2.
3. Determine the equation of the tangent line at point P.
- Use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the coordinate point and m is the slope.
- Substituting the values: y - 0.732 = 2(x - 60°).
4. Find the equation of the normal line at point P.
- The normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line's slope.
- The slope of the normal line = -1/2 (the negative reciprocal of the tangent slope).
- Using point-slope form: y - 0.732 = (-1/2)(x - 60°).
5. Find the x-intercepts of the tangent and normal lines.
- For the tangent line, set y = 0 and solve for x: 0 = 2(x - 60°).
- Solving for x gives x = 30° as the x-coordinate of the tangent line's x-intercept.
- For the normal line, set y = 0 and solve for x: 0 = (-1/2)(x - 60°).
- Solving for x gives x = 120° as the x-coordinate of the normal line's x-intercept.
6. Calculate the area of the quadrilateral.
- The quadrilateral is formed by the tangent line, the normal line, and the x-axis and y-axis.
- It can be divided into two triangles, each with a base on the x-axis and height to either the tangent line or normal line.
- Calculate the area of each triangle by using the formula: Area = 1/2 * base * height.
- Triangle 1 (with the tangent line): Area1 = 1/2 * (30° - 0°) * (0.732 - 0) = 11.92 square units.
- Triangle 2 (with the normal line): Area2 = 1/2 * (120° - 0°) * (0.732 - 0) = 43.92 square units.
- Add the areas of both triangles to determine the total area of the quadrilateral: Area = Area1 + Area2 = 11.92 + 43.92 = 55.84 square units.
Therefore, the area of the quadrilateral formed by the tangent, the normal at point P, and the coordinate axes is 55.84 square units.
it happens that the quadrilateral formed by those lines is a rectangle.
the tangent is horizontal and the normal is vertical.
y'(x) = 2cosx+2cos2x
y'(pi/3) = 1-1 = 0
y(pi/3) = √3+√3/2 = 3√3/2
so, we have the axes, and the lines y=3√3/2 and x=pi/3
the rectangle has area pi√3/2