# How to simplify velocity magnitude equations?

• Sep 29th 2009, 09:29 AM
Infernorage
How to simplify velocity magnitude equations?
In my calculus class a few days ago, my professor was going over how to find three dimensional velocity and its magnitude. In the problem, the magnitude of the velocity came out to be $\sqrt{9+36t^2+36t^4}=3+6t^2$.

My question is, how did he simplify the equation under the square root so that it became $3+6t^2$?

For the homework, I need to simplify these two square root equations to get something without a square root sign (if possible). Can you guys help me out with this? Thanks.
1) $\sqrt{9t^4+36+36t^2}$

2) $\sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$
• Sep 29th 2009, 09:52 AM
Rapha
Hello Infernorage!

Quote:

Originally Posted by Infernorage
In my calculus class a few days ago, my professor was going over how to find three dimensional velocity and its magnitude. In the problem, the magnitude of the velocity came out to be $\sqrt{9+36t^2+36t^4}=3+6t^2$.

Do you know this rule:

$(a+b)^2 = a^2+2ab+b^2$

It is not difficult to see that $9+36t^2+36t^4 = (6t^2+3)^2$

So

$\sqrt{9+36t^2+36t^4}=3+6t^2 = \sqrt{(6t^2+3)^2} = 3+6t^2$

Quote:

Originally Posted by Infernorage
My question is, how did he simplify the equation under the square root so that it became $3+6t^2$?

For the homework, I need to simplify these two square root equations to get something without a square root sign (if possible). Can you guys help me out with this? Thanks.
1) $\sqrt{9t^4+36+36t^2}$

This one is tricky.
I told you $(a+b)^2 = a^2+2ab+b^2$. You can't use it here unless :

$\sqrt{9t^4+36+36t^2} = \sqrt{9*(t^4+4+4t^2)}$

Hence

$\sqrt{9*(t^4+4+4t^2)} = \sqrt{9(t^2+2)^2}$

Quote:

Originally Posted by Infernorage
2) $\sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$

It is $sin^2+cos^2=1$

and furthermore

$\sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$

$=\sqrt{9cos^2t+16cos^2t+25sin^2t}$

$=\sqrt{25cos^2t+25sin^2t}$

$=\sqrt{25(cos^2t+sin^2t)}$

$=\sqrt{25*(1)} = \sqrt{25}$

Do you understand?

Yours
Rapha