Finding the range of a rational function

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  • #1
sooyong94
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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
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What did you try? Isn't it just some simple quadratic algebra?
 
  • #3
sooyong94
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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
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Some quartics look like quadratics, wouldn't you say?
 
  • #5
sooyong94
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Some quartics look like quadratics, wouldn't you say?
Yup, but how do I begin though?
 
  • #6
PeroK
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If ##(dy/dx)^2 < 9## what does that give you?
 
  • #7
sooyong94
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If ##(dy/dx)^2 < 9## what does that give you?
There are no values of t that satisfies the inequality?
 
  • #8
PeroK
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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
sooyong94
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Yes, but you've got to be prepared to do the algebra that shows why.
So how do I begin?
 
  • #10
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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?
 
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  • #11
sooyong94
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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
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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
sooyong94
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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
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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
sooyong94
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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
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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
sooyong94
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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
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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
sooyong94
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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
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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.
 

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