• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the range of a rational function

  • Thread starter sooyong94
  • Start date
  • #1
173
2

Homework Statement


A curve is given by the parametric equations
##x=t^2 +3##
##y=t(t^2+3)##
Find dy/dx in terms of t and show that (dy/dx)^2 >=9

Homework Equations


Parametric derivatives

The Attempt at a Solution


Using the chain rule, I arrived at ##\frac{dy}{dx}=\frac{3}{2}(\frac{t^2+1}{t})##
However, when I squared both sides to get ##(\frac{dy}{dx})^2##, I was unable to prove that (dy/dx)^2>=9. I know I need to find the range, however I'll need to graph the function, which proves to be tedious. Is there any workaround to this?
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,395
5,160
What did you try? Isn't it just some simple quadratic algebra?
 
  • #3
173
2
What did you try? Isn't it just some simple quadratic algebra?
I was thinking to graph the function, though I'll be dealing with quartic functions.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,395
5,160
Some quartics look like quadratics, wouldn't you say?
 
  • #5
173
2
Some quartics look like quadratics, wouldn't you say?
Yup, but how do I begin though?
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,395
5,160
If ##(dy/dx)^2 < 9## what does that give you?
 
  • #7
173
2
If ##(dy/dx)^2 < 9## what does that give you?
There are no values of t that satisfies the inequality?
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,395
5,160
There are no values of t that satisfies the inequality?
Yes, but you've got to be prepared to do the algebra that shows why.
 
  • #9
173
2
Yes, but you've got to be prepared to do the algebra that shows why.
So how do I begin?
 
  • #10
33,285
4,989
So how do I begin?
Your problem is equivalent to showing that ##(\frac{t^2 + 1}{t})^2 \ge 4##.
Presumably you are in a calculus class. How do you find the minimum value of a function?
 
Last edited:
  • #11
173
2
Your problem is equivalent to showing that ##(\frac{t^2 + 1}{t})^2 \ge 4##.
Presumably you are in a calculus class. How do you find the minimum value of a function?
Derive and set the value of the derivative to 0?
 
  • #12
33,285
4,989
Derive and set the value of the derivative to 0?
Yes, differentiate it and set the derivative to 0. But how will you know whether you have a minimum or maximum?

Now show us something...
 
  • #13
173
2
Yes, differentiate it and set the derivative to 0. But how will you know whether you have a minimum or maximum?

Now show us something...
But first, why did you say ((t^2+1)/t)^2 >4?

Use the first derivative test for critical points?
 
  • #14
33,285
4,989
But first, why did you say ((t^2+1)/t)^2 >4?
I changed what I wrote slightly, to ##(\frac{t^2 + 1}{t})^2 \ge 4##
sooyong94 said:
Use the first derivative test for critical points?
What do you think? Take a stab at it.
 
  • #15
173
2
I changed what I wrote slightly, to ##(\frac{t^2 + 1}{t})^2 \ge 4##

What do you think? Take a stab at it.
But what makes you think why ((t^2 +1)/t))^2 >=4?

The sign of the slopes changes?
 
  • #16
33,285
4,989
But what makes you think why ((t^2 +1)/t))^2 >=4?

The sign of the slopes changes?
In your first post you have
sooyong94 said:
Find dy/dx in terms of t and show that (dy/dx)^2 >=9
Isn't the inequality on the right equivalent to the one I wrote?
 
  • #17
173
2
In your first post you have

Isn't the inequality on the right equivalent to the one I wrote?
Wasn't it should be >=9?
 
  • #18
33,285
4,989
Wasn't it should be >=9?
You tell me.
  1. Start with (dy/dx)^2 >=9.
  2. Simplify.

This is post 18 in the thread. So far I haven't seen any work from you. What's stopping you?
 
  • #19
173
2
You tell me.
  1. Start with (dy/dx)^2 >=9.
  2. Simplify.

This is post 18 in the thread. So far I haven't seen any work from you. What's stopping you?
I have two inequalities, 3/2((t^2 +1)/t) <=-3 or ((t^2 +1)/t) >=2. Then how should I proceed from here?
 
  • #20
33,285
4,989
I have two inequalities, 3/2((t^2 +1)/t) <=-3 or ((t^2 +1)/t) >=2. Then how should I proceed from here?
##(dy/dx)^2 \ge 9##
##\Rightarrow (3/2)^2(\frac{t^2 + 1}{t})^2 \ge 9##
##\Rightarrow (\frac{t^2 + 1}{t})^2 \ge 4##
This last inequality is equivalent to ##\frac{t^2 + 1}{t} \ge 2## or ##\frac{t^2 + 1}{t} \le -2##
Since this is what you need to show, you can't assume that it's true. You can use calculus techniques to find the minimum and values of the rational expression.
 

Related Threads on Finding the range of a rational function

  • Last Post
Replies
6
Views
990
  • Last Post
Replies
2
Views
1K
Replies
3
Views
2K
Replies
11
Views
2K
Replies
14
Views
14K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
965
Replies
1
Views
1K
  • Last Post
Replies
20
Views
5K
Replies
8
Views
7K
Top