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
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.
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.
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)
Oops, I didn't calculate the intercepts. My bad.
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
Sure, I'd be happy to help you find the answers to your questions!
1) To find the angle at which the tangent line to the wave y = cos(x) crosses the x-axis at specific periods, let's first find the derivative of y = cos(x).
The derivative of cos(x) with respect to x is given by d/dx(cos(x)) = -sin(x).
Since the tangent line to a curve at a specific point represents the derivative of the curve at that point, we can set -sin(x) equal to the slope of the tangent line and solve for x.
In this case, the slope of the tangent line is 0 because it crosses the x-axis. So we have -sin(x) = 0.
Since sin(0) = 0, we find that x = 0. Therefore, the tangent line to the wave y = cos(x) at x = 0 crosses the x-axis at an angle of 0 degrees.
2) To find the area of an equilateral triangle ABC created by the x-axis and two tangent lines to the parabola y = 3 - x^2, we need to determine the coordinates of the vertices of the triangle.
Let's start by finding the coordinates of the points of tangency. The tangent lines will intersect the parabola at points where their slopes are equal to the derivative of the parabola.
The derivative of y = 3 - x^2 with respect to x is given by d/dx(3 - x^2) = -2x.
Setting -2x equal to the slopes of the tangent lines, we have -2x = m (where m is the slope). For an equilateral triangle, the slopes of the two tangent lines will be reciprocals of each other. So let's assume the first slope is m and the second slope would be -1/m.
Setting -2x = m and -2x = -1/m, we can solve for x to find the x-coordinates of the points of tangency.
For -2x = m: x = -m/2
For -2x = -1/m: x = 1/2m
Since these x-coordinates represent the same point of tangency, we can set them equal to each other and solve for m:
-m/2 = 1/2m
Cross multiplying, we have -m^2 = 1, which gives us m = ±√(-1).
Since the slope cannot be a complex number, there are no real values of m that satisfy this equation. Therefore, an equilateral triangle cannot be formed using the x-axis and two tangent lines to the parabola y = 3 - x^2.
I hope this helps! Let me know if you have any further questions.