# Finding the range of a rational function

## 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

PeroK
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What did you try? Isn't it just some simple quadratic algebra?

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.

PeroK
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Some quartics look like quadratics, wouldn't you say?

Some quartics look like quadratics, wouldn't you say?
Yup, but how do I begin though?

PeroK
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If ##(dy/dx)^2 < 9## what does that give you?

If ##(dy/dx)^2 < 9## what does that give you?
There are no values of t that satisfies the inequality?

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.

Yes, but you've got to be prepared to do the algebra that shows why.
So how do I begin?

Mark44
Mentor
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:
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?

Mark44
Mentor
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...

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?

Mark44
Mentor
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.

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?

Mark44
Mentor
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?

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?

Mark44
Mentor
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?

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?

Mark44
Mentor
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.