# Antiderivative of y

• WeiShan Ng

## Homework Statement

x=sec(t),
y=tan(t),
-π/2 ≤ t ≤ π/2
Try to find y in terms of x

## The Attempt at a Solution

1.[/B]
∂y/∂x = sec(t)/tan(t)
y=∫sec(t)/tan(t)∂x
=∫x/y∂x
=(1/y)*∫x∂x
=x2/2y + C
2y2=x2 + C
When t=π/4, x=√2, y=1
2(1)2 = (√2)2 + C
C=0
So y2 = x2/2

2.
y/x = sin(t)
1/x = cos(t)
sin2(t) + cos2(t) = 1
y2/x2 + 1/x2 = 1
y2 = x2 + 1

Why couldn't I get the same equations in both (1) and (2)? It turns out only the equation 2 works well with other value of t. What did I do wrong in (1)?

## Homework Statement

x=sec(t),
y=tan(t),
-π/2 ≤ t ≤ π/2
Try to find y in terms of x
What does this have to do with antiderivatives, as stated in your thread title?
WeiShan Ng said:

## The Attempt at a Solution

1.[/B]
∂y/∂x = sec(t)/tan(t)
No -- this is just y/x, not the partial of y with respect to x. Finding y in terms of x can be done using only some trig identities and some algebra.
This part makes no sense.
WeiShan Ng said:
y=∫sec(t)/tan(t)∂x
=∫x/y∂x
=(1/y)*∫x∂x
=x2/2y + C
2y2=x2 + C
When t=π/4, x=√2, y=1
2(1)2 = (√2)2 + C
C=0
So y2 = x2/2

2.
This part does make sense, but you have some errors.
WeiShan Ng said:
y/x = sin(t)
1/x = cos(t)
sin2(t) + cos2(t) = 1
y2/x2 + 1/x2 = 1
OK to here, but you have a mistake in the work below.
WeiShan Ng said:
y2 = x2 + 1

Why couldn't I get the same equations in both (1) and (2)? It turns out only the equation 2 works well with other value of t. What did I do wrong in (1)?

Last edited:
You've treated ##y## as a constant with respect to ##dx##. But as both are functions of ##t## and your are looking for ##y(x)##. How can this be?

What does this have to do with antiderivatives, as stated in your thread title?
No -- this is just y/x, not the partial of y with respect to x. Finding y in terms of x can be done using only some trig identities and some algebra.
This part makes no sense.
This part does make sense, but you have some errors.
OK to here, but you have a mistake in the work below.

1. Ooops. Sorry. I actually don't quite understand the meaning of antiderivative...><
2. I thought since ∂y/∂x = (∂y/∂t)/(∂x/∂t), and ∂y/∂t = sec2 (t), ∂x/∂t = sec(t)tan(t), then ∂y/∂x = sec(t)/tan(t) ??
3. Ooops. It's y2 = x2 - 1

Thanks for the help!

You've treated ##y## as a constant with respect to ##dx##. But as both are functions of ##t## and your are looking for ##y(x)##. How can this be?
I thought when we integrate a function with respect to only one of its variables, we treat the rest of the variables as constants?

I thought when we integrate a function with respect to only one of its variables, we treat the rest of the variables as constants?

I can't see the relevance of that here.

I thought when we integrate a function with respect to only one of its variables, we treat the rest of the variables as constants?
There is no need to use any calculus in answering the question.

WeiShan Ng said:
∂y/∂x = sec(t)/tan(t)

2. I thought since ∂y/∂x = (∂y/∂t)/(∂x/∂t), and ∂y/∂t = sec2 (t), ∂x/∂t = sec(t)tan(t), then ∂y/∂x = sec(t)/tan(t) ??
All of your derivatives you show as being partial derivatives (e.g.,∂y/∂x) . Since both x and y are functions of a single variable, t, ordinary derivatives are appropriate here. With x = sec(t), dx/dt = sec(t) tan(t), and similar for y and dy/dt.

In any case, as already stated by haruspex, you don't need any calculus for this problem.

All of your derivatives you show as being partial derivatives (e.g.,∂y/∂x) . Since both x and y are functions of a single variable, t, ordinary derivatives are appropriate here. With x = sec(t), dx/dt = sec(t) tan(t), and similar for y and dy/dt.

In any case, as already stated by haruspex, you don't need any calculus for this problem.
Thanks for pointing it out. I thought they are interchangeable until now...Paradigm shift! Now I finally understand why my lecturer uses them .

So I can't solve it using Calculus?

So I can't solve it using Calculus?

You could if you wanted, but it's unnecessarily complicated. In this case, if you know your trig identiies, you are essentially given:

##y^2 = x^2 - 1##

So, why use calculus to solve that?

So I can't solve it using Calculus?
The problem as stated:

## Homework Statement

x=sec(t),
y=tan(t),
-π/2 ≤ t ≤ π/2
Try to find y in terms of x
To reiterate what has already been stated (by @PeroK), this problem has nothing to do with calculus.

PeroK said:
So, why use calculus to solve that?
I know I can do it using trig identities but I was just curious whether the question can be solved using calculus. I tried and failed and don't know where did I get wrong.

Last edited by a moderator:
So, why use calculus to solve that? know I can do it using trig identities but I was just curious whether the question can be solved using calculus. I tried and failed and don't know where did I get wrong.

Can you tell which formulas you used?

Why is ##\frac{dy}{dx}=\frac{\sec t}{\tan t} ##? (I think it's correct but don't have it in mind.)

And why do you think you can treat ##y## as a constant with respect to the integration over ##x##?
Isn't ##y=y(x)=\tan (arc\sec x))## dependent on ##x##?

So, why use calculus to solve that?
I know I can do it using trig identities but I was just curious whether the question can be solved using calculus. I tried and failed and don't know where did I get wrong.
Really, the question is "What's the right tool for the job?"

The original problem is to eliminate the parameter t in two equations with trig functions, so all that is required is a bit of algebra and a trig identity or two. The problem doesn't have anything to do with rates of change or summing infinitesimally small quantities, so calculus methods are neither called for or needed.

Really, the question is "What's the right tool for the job?"

The original problem is to eliminate the parameter t in two equations with trig functions, so all that is required is a bit of algebra and a trig identity or two. The problem doesn't have anything to do with rates of change or summing infinitesimally small quantities, so calculus methods are neither called for or needed.
I don't think it can be solved using Galois Theory either.