The graph of y=cos x * ln cos^2x has seven horizontal tangent lines on the interval [0,2pi]. Find the x-coordinate of all points at which these tangent lines occur.
12 years ago
12 years ago
first of all re-write it as
y = cosx ( 2 ln(cosx) ) using log rules
= 2 cosx ln(cosx)
now use the product rule to find
dy/dx = 2cosx (-sinx/cosx) + (-2sinx)(ln(cosx))
= - 2sinx(1 + ln(cosx) )
= 0 for horizontal tangents
-2sinx = 0 or 1 + ln(cosx) = 0
for -2sinx = 0
sinx = 0
x = 0, π , 2π
for 1 + ln(cox) = 0
ln(cosx) = -1
cosx = e^-1 = 1/e
x = 1.194 or 2Ï€ - 1.194
I only found 5 values, don't know how they got the 7 values.
I was going to replace cos^2 x with (cos 2x + 1)/2
but ran into an awful mess trying to solve the derivative for zero
1 year ago
To find the x-coordinate of all points at which the graph of y = cos(x) * ln(cos^2(x)) has horizontal tangent lines on the interval [0, 2Ï€], we need to find the values of x where the derivative of the function is equal to zero.
1. First, let's find the derivative of the function:
The derivative of ln(u) is du/u, so we need to find the derivative of the inside function cos^2(x)
Using the chain rule, the derivative of cos^2(x) is -2cos(x)sin(x).
The derivative of y with respect to x is:
dy/dx = -2cos^2(x)sin(x) + ln(cos^2(x)) * -sin(x)
2. Next, let's set the derivative equal to zero and solve for x:
-2cos^2(x)sin(x) + ln(cos^2(x)) * -sin(x) = 0
Factoring out sin(x), we get:
sin(x) * (-2cos^2(x) + ln(cos^2(x))) = 0
3. Since sin(x) = 0 has infinitely many solutions on the interval [0, 2Ï€], we need to solve the equation
-2cos^2(x) + ln(cos^2(x)) = 0
4. Let's solve this equation numerically using a graphing calculator or software, or we can use an iterative method:
-2cos^2(x) + ln(cos^2(x)) = 0
-2cos^2(x) = ln(cos^2(x))
e^(-2cos^2(x)) = cos^2(x)
e^(-2cos^2(x)) - cos^2(x) = 0
By using an iterative method, we can approximate the solutions:
One solution is x ≈ 1.06419 (approximately)
Another solution is x ≈ 1.47608 (approximately)
And so on...
5. Continue the iterative process or use a graphing calculator to find more solutions until you have found seven distinct solutions within the interval [0, 2Ï€].
Therefore, the x-coordinates of the points where the graph has horizontal tangent lines are approximately 1.06419, 1.47608, and so on, until a total of seven solutions are found.
1 year ago
To find the x-coordinate of the points where the tangent lines occur, we need to find the derivative of the function y = cos(x) * ln(cos^2(x)), and then set it equal to zero to find the points where the slope is zero (i.e., horizontal tangent lines).
Let's start by finding the derivative of y with respect to x using the product rule and chain rule.
Given: y = cos(x) * ln(cos^2(x))
Using the product rule:
dy/dx = [d(cos(x))/dx * ln(cos^2(x))] + [cos(x) * d(ln(cos^2(x)))/dx]
The derivative of cos(x) with respect to x is -sin(x).
dy/dx = [-sin(x) * ln(cos^2(x))] + [cos(x) * d(ln(cos^2(x)))/dx]
Next, let's find the derivative of ln(cos^2(x)) using the chain rule.
Let u = cos^2(x)
Then ln(u) = ln(cos^2(x))
Using the chain rule:
d(ln(u))/dx = (1/u) * du/dx
du/dx = d(cos^2(x))/dx = 2cos(x) * (-sin(x)) = -2cos(x)sin(x)
Therefore, d(ln(u))/dx = (1/u) * (-2cos(x)sin(x))
Substituting back into the derivative equation:
dy/dx = [-sin(x) * ln(cos^2(x))] + [cos(x) * (1/cos^2(x)) * (-2cos(x)sin(x))]
= [-sin(x) * ln(cos^2(x))] - [2sin(x)cos(x)]
= -sin(x) * [ln(cos^2(x)) + 2cos(x)]
To find the points where the tangent lines occur, we need to set the derivative equal to zero and solve for x:
dy/dx = -sin(x) * [ln(cos^2(x)) + 2cos(x)] = 0
Since sin(x) cannot be equal to zero for any value of x in the interval [0, 2Ï€], we need to set the expression inside the square brackets equal to zero:
ln(cos^2(x)) + 2cos(x) = 0
Now, we can solve this equation for x:
ln(cos^2(x)) + 2cos(x) = 0
First, let's solve for cos^2(x):
cos^2(x) = e^(-2cos(x))
Taking the square root of both sides:
cos(x) = ±sqrt(e^(-2cos(x)))
Since cos(x) cannot be negative in the given interval [0, 2Ï€], we only consider the positive square root:
cos(x) = sqrt(e^(-2cos(x)))
Taking the natural logarithm of both sides:
ln(cos(x)) = -cos(x)
Now we have an equation in terms of cos(x) that we can solve algebraically. However, solving this equation analytically may not be straightforward. To find the x-values where this equation is satisfied, we can use numerical methods such as graphing calculators, numerical approximation methods, or computer software.
Using a graphing calculator or graphing software, we can plot the graph of y = cos(x) * ln(cos^2(x)) and find the x-coordinate of the points where the graph intersects the x-axis, corresponding to the horizontal tangent lines. The points where the graph crosses the x-axis are the x-coordinates of the tangent points.