Calculus problem?

Question

Find the points on the graph of y=sec x , 0>(or equal to)x>(or equal to)2pi, where the tangent is parallel to the linde 3y-2x=4

Answer 1

Rewrite the line in slope-intercept form:
y=2/3 x + 4/3
So you're looking for points on the graph of y=sec x such that dy/dx = 2/3
dy/dx= sin x/cos² x = 2/3
sin x/(1-sin² x) = 2/3
sin x=2/3 - 2/3 sin² x
2/3 sin² x + sin x=2/3
sin² x + 3/2 sin x = 1
sin² x + 3/2 sin x + 9/16 = 25/16
(sin x + 3/4)² = 25/16
sin x + 3/4 = ±5/4
sin x = -3/4 ±5/4
sin x = -2 or sin x = 1/2

Clearly, sin x can't equal -2, so you're looking for points where sin x = 1/2. There are two such points - π/6 and 5π/6 (i.e. 30° and 150° - you should remember them from your special right triangles). Therefore, your solution set is:

{π/6, 5π/6}

Answer 2

You want the points on the y = sec x graph to be where the tangent line is parallel to 3Y-2x=4. I used lower case and upper case for the two functions. Take dy/dx (y = 4/3+(2/3)x, dy/dx = 2/3)and dY/dx (you can look this one up) and set them equal to each other. You are basically finding the points where the two lines have the same slope. Solve and find the x's between 0 and 2pi that satisfy your first derivative equation.

Plug the x (or x's) into the sec x graph to get the corresponding y values.

Answer 3

The slope of the tangent = slope of the given line = 2/3
Also, slope of the tangent = y' = sec x tanx

>>
sec x tan x = 2/3
>>
sinx / cos^2 x = 2/3
sinx / ( 1 - sin^2 x) = 2/3
>>
sinx = 0.5
>> x = pi/12 or 11pi/12