Question

Hello,
I have two questions that I've been working on, but I don't know how to get the answer...something similar is going to be on the test so I hope I can get some help, thanks!


1) At what angle does the tangent line to the wave y=cosx at periods (pi/3, 1/2) and (-pi/3, 1/2) cross x-axis.


2) Find the area of an equilateral tiangle ABC created by x-axis and 2 tangent lines to the parabola y= 3-x^2

Answer 1

1) The coordinates you have provided, where you want to know the slope, are not "periods". They are locations on the y vs. x "wave" curve.
The slope of the tangent line is the same thing as the derivative, dy/dx. For this y(x) function, dy/dx = -sinx. The slope is (+sqrt 3)/2 for the line that is tangent at x=pi/3, and -(sqrt3)/2 for the line that is tangent at x = -pi/3.

The slope remains the same where it crosses the x axis. The angle with a tangent of sqrt3/2 is 40.89 degrees (0.7137 radians)

2) You have to specify where the tangents to the parabola are located. If the lines are drawn tangent where the parabola intercepts the x-axis, at y=0, then those triangle corners are at x = +sqrt3 and -sqrt3. The base of the triangle then has a length of 2 sqrt3. Assume the triangle is equilateral when computing the area, since they say it is.

Answer 2

I wonder if the second question isn't a little more mean and evil than that? :-)

The parabola is upturnd, symmetrical around x=0, like a rounded hill.

A tangent to that parabola that forms a side of an equilateral triangle as described would be of the form +/-sqrt(3)x + c (because of the 60 degrees needed for the equilateral) and would intersect the parabola at exactly one point. A line drawn tangent to the intercept willnot be at 60 degrees, I think.

Answer 3

Since the triangle is equilateral, the angles must be 60º
dy/dx for the function is -2x
tan 60º = √3 and the slope of the two tangents must then be -√3 in the first quadrant, and +√3 in the second quadrant.
-2x = -√3, so x = √3/2
then y = 3 - (√3/2)^2 = 9/4
so the point of contact of the tangent in quadrant I is (√3/2,-/4) and slope is -√3

I found the equation of that tangent to be y = -√3x + 15/4

so the height of the triangle is 15/4, the y-intercept, and the base of the triangle is 15/(2√3), twice the x-intercept
thus the area
= (1/2)(15/(2√3))(15/4)
= 225/(16√3)

Answer 4

Oops, I didn't calculate the intercepts. My bad.

Answer 5

Equilateral triangle symmetrical about the y-axis means that the two sides which are the tangents have gradients
-√3 and +√3.
y' = -2x
So the points of contact of the tangent to the parabola are (-√3/2, 9/4) and (√3/2, 9/4).
Hence we can calculate the x and the y intercepts of the tangents to the parabola.

y-int is (9+2√3)/4. x-int -(2+3√3)/4 and (2+3√3)/4.

Hence the required area is
(9+2√3)(2+3√3)/16 = (36+31√3)/16