# Finding the range of a rational function

sooyong94

## 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?

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

sooyong94
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.

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

sooyong94
Some quartics look like quadratics, wouldn't you say?
Yup, but how do I begin though?

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

sooyong94
If ##(dy/dx)^2 < 9## what does that give you?
There are no values of t that satisfies the inequality?

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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.

sooyong94
Yes, but you've got to be prepared to do the algebra that shows why.
So how do I begin?

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:
sooyong94
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?

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...

sooyong94
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?

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.

sooyong94
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?

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?

sooyong94
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?

Mentor
Wasn't it should be >=9?
You tell me.
2. Simplify.

This is post 18 in the thread. So far I haven't seen any work from you. What's stopping you?

sooyong94
You tell me.