Proving Horizontal Tangents of y=Cos(x) & y=Sec(x) at x=0

In summary: This means that at x = 0, the derivative is indeed 0, proving that the function has a horizontal tangent at that point.In summary, the conversation was about proving the functions y=Cos(x) and y=Sec(x) have horizontal tangents at x=0. The person was able to prove the first function, but was unsure of how to prove the second one. They were trying to show that Sec(x)Tan(x)=0, but were confused about how to work with 1/cos(x) and how to divide by 0 to get 0. However, they later realized that their initial approach was correct and that at x=0,
  • #1
Robokapp
218
0
im supposed to prove that

y=Cos(x)

and

y=Sec(x)

have horisontal tangents for x=0

i got the derivatives just fine, i proved the first on e with no problem, but what do i do with second? i get to a point where i must prove that Sec(x)Tan(x)=0 but sec x = 1/cos(x) and i can't work with that...i mean how can i divide 1 by something to get zero? it's an asymptote...so what do i do?

i know how to get derivatives, i don't know how to prove that the second one is having the horisontal tangent.
 
Physics news on Phys.org
  • #2
y = secx
y' = secx.tanx = sinx/cos²x

at x = 0, sinx = 0, cosx = 1

y' = 0/1² = 0
==========
 
  • #3
Fermat said:
y = secx
y' = secx.tanx = sinx/cos²x

at x = 0, sinx = 0, cosx = 1

y' = 0/1² = 0
==========

okay...i thought that when you have secx*tanx=0 you have to set them like when you have parenthesis...each one at a time. That was probably a better explanation than most teachers would give. Thank you.

What i was trying to do is prove that either tan or sec is some value and the second one is zero, so zero*value=0 but it doesn't work that way.

can i ask though...why was i wrong? i mean if you do (x)(X+1)=0 you're defenetly not wrong to set either one equl to zero...why wasn't my way working? (To be honest I am more interested on what went wrong than what the answer is...i can always copy that from the book but i don't want to).
 
  • #4
As far as I can see, you weren't wrong!

You had,

y' = secx*tanx

at x = 0, secx= 1/cosx = 1/1 = 1, and tanx = 0,

So,

y' = 1*0 = 0

Waht you have is secx = 1 and the other one, tanx = 0
 

FAQ: Proving Horizontal Tangents of y=Cos(x) & y=Sec(x) at x=0

How do you prove that the function y = Cos(x) has a horizontal tangent at x = 0?

To prove that the function y = Cos(x) has a horizontal tangent at x = 0, we can use the definition of a horizontal tangent, which states that the slope of the tangent line at a point must be equal to 0. We can find the slope of the tangent line by taking the derivative of the function, which in this case is y' = -Sin(x). When x = 0, the derivative is also 0, indicating a horizontal tangent.

Can you explain the concept of a horizontal tangent in relation to the function y = Cos(x)?

A horizontal tangent for the function y = Cos(x) means that at the point x = 0, the slope of the tangent line is 0. This means that the tangent line is parallel to the x-axis and has no slope, resulting in a horizontal line. In other words, at x = 0, the rate of change of the function is 0, indicating a flat or constant value.

How can you prove that the function y = Sec(x) has a horizontal tangent at x = 0?

To prove that the function y = Sec(x) has a horizontal tangent at x = 0, we can follow the same process as for the function y = Cos(x). We take the derivative of the function, which is y' = Sec(x)Tan(x), and when x = 0, the derivative is 0, indicating a horizontal tangent. Additionally, we can also use the fact that the derivative of the Secant function is equal to the Secant function multiplied by the Tangent function, which also evaluates to 0 at x = 0.

What is the significance of proving horizontal tangents for the functions y = Cos(x) and y = Sec(x) at x = 0?

Proving the existence of horizontal tangents for these functions at x = 0 is important because it shows that the functions have a constant value at that point. This can be useful in various applications, such as calculating the area under the curve or finding the average rate of change of the function. It also helps us understand the behavior of the functions at that specific point.

Are there any other methods for proving horizontal tangents of y = Cos(x) & y = Sec(x) at x = 0?

Yes, there are other methods for proving horizontal tangents for these functions at x = 0. One method is to use the Maclaurin series expansion for Cos(x) and Sec(x), which shows that the functions have a constant value at x = 0. Another approach is to use the limit definition of the derivative, which also evaluates to 0 at x = 0 for both functions. These methods provide alternative ways to prove the existence of horizontal tangents at a given point.

Similar threads

Differentiate ##f(x)=x\cos{x}+2\tan{x}: D/dx ##\tiny{2.4.2}##
Replies
10
Views
1K
Calculus ## G(y, z)=z\frac{\partial}{\partial z}F(y, z)-F(y, z) ##?
Replies
6
Views
1K
Solve the given first order differential equation
Replies
8
Views
1K
Quadratics Having a Common Tangent
Replies
1
Views
646
##\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)##
Replies
5
Views
158
Lagrange multipliers understanding
Replies
8
Views
1K
Using Euler's formula to prove trig identities using "sum to product" technique
Replies
4
Views
1K
Find Values of Osculating Circle Tangent to a Parabola
Replies
4
Views
842
Proof given ##x < y < z## and a twice differentiable function
Replies
8
Views
997
Prove if ##x<0## and ##y<z## then ##xy>xz## (Rudin)
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top