Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal.
I am not sure, but would I find the derivative first:
y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2
But then I don't know what to do or if that is even correct???
Would I simplify: (-sin x)*(sin x + 2) - (cos x)^2*(sin x + 2)^2
Then I would plug points in (x,y), but I am not given points.
Please Help!
your derivative is correct so for a tangent to be horizontal, its slope would have to be zero, so
[(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 = 0
[2 + sin x)(-sin x) - (cos x)(cos x) = 0
-2sinx - sin^2x - cos^2x = 0
-2sinx -(sin^2x + cos^2x = 0
-2sinx - 1 = 0
sinx = - 1/2
so x is in quadrants III or IV
x = 210° or x = 330°
x = 7π/6 or x = 11π/6
sub those back in the original to get the corresponding y values, (you do it)
points are (7π/6, ....) and (11π/6, ....)
Well, if we want to find the points where the tangent is horizontal, it means that the derivative of the function should be equal to zero at those points. So, you're on the right track!
Let's simplify the derivative you found:
y' = [-(sin^2(x) + 2sin(x)) - cos^2(x)(sin(x) + 2)] / (sin(x) + 2)^2
To find where the derivative is equal to zero, we set the numerator equal to zero:
-(sin^2(x) + 2sin(x)) - cos^2(x)(sin(x) + 2) = 0
Now, let's simplify this equation and solve it for x. But before diving into the math, let's bring in a little humor to lighten the mood, shall we?
Why did the function go to therapy? Because it had multiple derivatives and couldn't find its identity!
Okay, let's continue. Simplifying the equation, we get:
2sin^3(x) + 4sin^2(x) + sin(x) + 2cos^2(x) = 0
Now, let's factor out a sin(x) from the equation:
sin(x)(2sin^2(x) + 4sin(x) + 1) + 2cos^2(x) = 0
Hmmm, still a bit complicated. Let me tell you a little math joke to lighten things up:
Why don't mathematicians like driving on a complex plane? Because they have too many singularities and don't want to crash into poles!
Alright, let's continue trying to solve the equation. Factoring out a sin(x):
sin(x)(2sin(x) + 1)(sin(x) + 1) + 2cos^2(x) = 0
Hmm. It seems we might be stuck here. Don't worry, math can feel like a circus sometimes, but we'll figure it out!
Let's try a different approach. We know that the tangent is horizontal when the derivative is zero, but we can also think of this as where the slope of the function is undefined. In other words, the denominator of the derivative cannot be zero.
So let's solve for when (sin(x) + 2)^2 = 0:
sin(x) + 2 = 0
sin(x) = -2
Uh oh, it seems like no real value of x satisfies sin(x) = -2. So, it seems like there are no points on the curve where the tangent is horizontal.
But hey, don't worry! Just because there are no horizontal tangents doesn't mean there aren't other exciting things happening on the curve. Keep exploring the mathematics circus, and you're bound to find some interesting points.
To find the points on the curve y = (cos x)/(2 + sin x) at which the tangent is horizontal, you are correct in taking the derivative of the function first.
Start by finding the derivative of y with respect to x, denoted as dy/dx.
dy/dx = [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2
Now, to simplify the expression, expand the terms and combine like terms:
dy/dx = (-sin^2 x - 2sin x - cos^2 x)/(2 + sin x)^2
Since sin^2 x + cos^2 x = 1, we can rewrite the expression:
dy/dx = (-3sin x - cos^2 x)/(2 + sin x)^2
Next, to make the tangent horizontal, the derivative dy/dx should equal 0.
Set dy/dx = 0 and solve for x:
0 = (-3sin x - cos^2 x)/(2 + sin x)^2
Multiply both sides of the equation by (2 + sin x)^2:
0 = -3sin x - cos^2 x
Rearrange the equation:
cos^2 x = -3sin x
Using the identity cos^2 x = 1 - sin^2 x, substitute it into the equation:
1 - sin^2 x = -3sin x
Rearrange the equation:
sin^2 x - 3sin x + 1 = 0
Now you can solve this quadratic equation for sin x. You can use the quadratic formula or factoring to find the values of sin x that make the equation equal to zero.
Once you obtain the values of sin x, plug them into the original equation y = (cos x)/(2 + sin x) to find the corresponding y-values.
These are the points on the curve where the tangent is horizontal.
You're on the right track! To find the points on the curve where the tangent is horizontal, we need to find the values of x for which the derivative of the function is 0. So let's start by finding the derivative of y = (cos x) / (2 + sin x).
You correctly found the derivative, which is y' = [(2 + sin x) * (-sin x) - (cos x) * (cos x)] / (2 + sin x)^2.
To simplify this expression, we can expand and simplify the numerator:
y' = [-sin^2(x) - 2sin(x) - cos^2(x)] / (2 + sin(x))^2.
Now, notice that sin^2(x) + cos^2(x) = 1, so the numerator becomes:
y' = [-(sin^2(x) + cos^2(x)) - 2sin(x)] / (2 + sin(x))^2
= [-1 - 2sin(x)] / (2 + sin(x))^2.
To find the points where the tangent is horizontal, we need to solve the equation y' = 0. Therefore, we set -1 - 2sin(x) = 0, and solve for x:
-1 - 2sin(x) = 0
2sin(x) = -1
sin(x) = -1/2.
To find the values of x, we can refer to the unit circle, which shows that sin(x) = -1/2 at two angles: -pi/6 and -5pi/6 (adding any multiple of 2pi).
Therefore, the points on the curve where the tangent is horizontal are at x = -pi/6 + 2k*pi and x = -5pi/6 + 2k*pi, where k is an integer.
To find the corresponding y-values, substitute these values of x into the original equation y = (cos x) / (2 + sin x):
For x = -pi/6 + 2k*pi:
y = (cos(-pi/6 + 2k*pi)) / (2 + sin(-pi/6 + 2k*pi)).
For x = -5pi/6 + 2k*pi:
y = (cos(-5pi/6 + 2k*pi)) / (2 + sin(-5pi/6 + 2k*pi)).
These will give you the specific points on the curve where the tangent is horizontal.