Find an equation for the tangent line to the curve at (π/2 , 2).
y = 4 + cot(x) - 2csc(x)
I am confused how to take the derivative of this problem. When I tried to solve it I ended up with
-csc^2 (x) + (2csc(x) * cot(x)).
From there I can't seem to simplify it.
Also, can you explain how to find the horizontal tangent.
your derivative is correct.
now just plug in x = π/2 to get the slope
remember that tan(π/2) is undefined, BUT
cot (π/2) = cos(π/2) / sin(π/2) = 0/1 = 0
so slope = -csc^2 (π/2) = 2
equation of tangent:
y-1 = 2(x- π/2)
clean it up if needed.
for a horizontal tangent, remember that a horizontal line has a slope of 0.
I calculated and got -1 as the slope. Can you please explain how you got 2 for the slope?
To find the equation of the tangent line to the curve at (π/2, 2), we need to find the derivative of the curve and evaluate it at x = π/2.
First, let's find the derivative of the function y = 4 + cot(x) - 2csc(x).
The derivative of cot(x) is -csc^2(x), and the derivative of -2csc(x) is 2csc(x) * cot(x). The derivative of a constant (like 4) is zero. Thus, the derivative of y with respect to x is:
dy/dx = -csc^2(x) + 2csc(x) * cot(x).
Now, let's evaluate the derivative at x = π/2 to find the slope of the tangent line at that point:
dy/dx = -csc^2(π/2) + 2csc(π/2) * cot(π/2).
We know that csc(π/2) = 1 and cot(π/2) = 0. Therefore, the slope of the tangent line at x = π/2 is:
dy/dx = -csc^2(π/2) + 2csc(π/2) * cot(π/2) = -1 + 2 * 1 * 0 = -1.
To find the equation of the tangent line, we will use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the curve (in this case, (π/2, 2)) and m is the slope of the tangent line.
Using the point (π/2, 2) and the slope -1, we can plug these values into the point-slope form:
y - 2 = -1(x - π/2).
To simplify, distribute the -1 on the right side:
y - 2 = -x + π/2.
Now, we rearrange the equation to solve for y:
y = -x + π/2 + 2.
Combining like terms, we have:
y = -x + π/2 + 4/2.
Simplifying further, we get:
y = -x + π/2 + 2.
So, the equation of the tangent line to the curve at (π/2, 2) is y = -x + π/2 + 2.
Now, let's discuss how to find the horizontal tangent.
A horizontal tangent occurs when the slope of the curve is zero. So, to find horizontal tangents, we need to set the derivative equal to zero and solve for x.
In this case, we have:
dy/dx = -csc^2(x) + 2csc(x) * cot(x) = 0.
To simplify this equation, we can factor out a common factor:
0 = csc(x)(2cot(x) - csc(x)).
Now, we can set each factor equal to zero:
1) csc(x) = 0.
2) 2cot(x) - csc(x) = 0.
To solve equation 1), we know that csc(x) = 1/sin(x), so 1/sin(x) = 0 has no solutions. Therefore, csc(x) = 0 has no solutions.
To solve equation 2), we can rearrange it:
2cot(x) = csc(x).
Using the identity for cotangent (cot(x) = cos(x)/sin(x)) and cosecant (csc(x) = 1/sin(x)), we have:
2cos(x)/sin(x) = 1/sin(x).
Now, cross-multiplying yields:
2cos(x) = 1.
Dividing both sides by 2, we get:
cos(x) = 1/2.
From here, we can use inverse cosine to find the values of x. The solutions to cos(x) = 1/2 are x = π/3 and x = 5π/3.
Therefore, the curve has horizontal tangents at x = π/3 and x = 5π/3.
To find the equation of the tangent line to the curve at a specific point (π/2 , 2), you need to find the derivative of the function, plug in the x-value of the point to find the slope, and then use the point-slope formula to find the equation of the tangent line.
Let's start with finding the derivative of the function y = 4 + cot(x) - 2csc(x).
To find the derivative, you can use the quotient rule and chain rule. The derivative of cot(x) is -csc^2(x), and the derivative of csc(x) is -csc(x) * cot(x).
Applying the quotient rule and chain rule, the derivative of y with respect to x is:
dy/dx = -csc^2(x) - 2[-csc(x) * cot(x)]
Now, let's evaluate the derivative at x = π/2, which is the x-value of the point where the tangent line is desired.
dy/dx = -csc^2(π/2) - 2[-csc(π/2) * cot(π/2)]
Since cot(π/2) = 0 and csc(π/2) = 1, the derivative becomes:
dy/dx = -csc^2(π/2) - 2[-csc(π/2) * 0]
dy/dx = -csc^2(π/2) - 2[0]
dy/dx = -csc^2(π/2)
Now, to simplify further, recall that csc^2(x) = 1/sin^2(x).
Therefore, dy/dx = -1/sin^2(π/2).
Now, sin(π/2) = 1, so sin^2(π/2) = 1^2 = 1.
Substituting back into the equation, dy/dx = -1/1 = -1.
Now that we have the slope of the tangent line, which is -1, we can use the point-slope formula to find the equation of the line.
Using the point (π/2 , 2), the equation of the tangent line is:
y - 2 = -1(x - π/2)
y - 2 = -x + π/2
Rearranging the equation, we get:
y = -x + π/2 + 2
Simplifying further:
y = -x + π/2 + 4/2
y = -x + π/2 + 2
Therefore, the equation of the tangent line to the curve at (π/2 , 2) is y = -x + π/2 + 2.
Now, as for finding the horizontal tangent, you need to find the points where the derivative is zero. In this case, since we already found the derivative to be -csc^2(x), the horizontal tangent occurs when -csc^2(x) = 0.
However, the csc^2(x) is always positive or zero, and it can never be equal to zero. Therefore, there are no Horizontal Tangents for the given function.