# analysis vector

• Jun 17th 2008, 09:18 AM
noob
analysis vector
could anyone please tell me how to differentiate those equation;

r(t) = ln (t^2 + 1)i + (t - 2 tan^-1 t)j + 2 (surd2) k

i note that to differentiate ln (t^2 + 1)i = 1/t^2 +1 . (2t) how bout j?
• Jun 17th 2008, 09:26 AM
Soroban
Hello, noob!

Quote:

Could anyone please tell me how to differentiate this equation?

. . $\displaystyle r(t) \:= \:\ln(t^2 + 1)i + (t - 2\tan^{-1}\!t)j + 2\sqrt{2}k$

$\displaystyle f(t)\:=\:\ln(t^2+1) \quad\Rightarrow\quad f'(x) \:=\:\frac{2t}{t^2+1}$

$\displaystyle g(t) \:=\:t - 2\tan^{-1}\!t \quad\Rightarrow\quad g'(t) \:=\:1 - \frac{2}{1 + t^2}$

$\displaystyle h(x) \:=\:2\sqrt{2} \quad\Rightarrow\quad h'(x) \:=\:0$

• Jun 17th 2008, 09:35 AM
noob
Quote:

Originally Posted by Soroban
Hello, noob!

$\displaystyle f(t)\:=\:\ln(t^2+1) \quad\Rightarrow\quad f'(x) \:=\:\frac{2t}{t^2+1}$

$\displaystyle g(t) \:=\:t - 2\tan^{-1}\!t \quad\Rightarrow\quad g'(t) \:=\:1 - \frac{2}{1 + t^2}$

$\displaystyle h(x) \:=\:2\sqrt{2} \quad\Rightarrow\quad h'(x) \:=\:0$

im just confusing on how does the g(t) is obtain, maybe got some formula did i forgot
• Jun 17th 2008, 09:39 AM
TheEmptySet
Quote:

Originally Posted by noob
im just confusing on how does the g(t) is obtain, maybe got some formula did i forgot

$\displaystyle y=tan^{-1}(x) \iff \tan(y)=x$

Now taking the implict derivative with respect to x we get

$\displaystyle sec^2(y)\cdot \frac{dy}{dx}=1 \iff \frac{dy}{dx}=\frac{1}{\sec^2(y)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^ 2(y)}=\frac{1}{1+x^2}$

The last step comes from the eqaution at the top $\displaystyle \tan(y)=x$
• Jun 17th 2008, 09:51 AM
noob
Quote:

Originally Posted by TheEmptySet

$\displaystyle y=tan^{-1}(x) \iff \tan(y)=x$

Now taking the implict derivative with respect to x we get

$\displaystyle sec^2(y)\cdot \frac{dy}{dx}=1 \iff \frac{dy}{dx}=\frac{1}{\sec^2(y)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^ 2(y)}=\frac{1}{1+x^2}$

The last step comes from the eqaution at the top $\displaystyle \tan(y)=x$

i was really help mates, i now clearly understand how it comes
• Jun 17th 2008, 09:52 AM
noob
Quote:

Originally Posted by TheEmptySet

$\displaystyle y=tan^{-1}(x) \iff \tan(y)=x$

Now taking the implict derivative with respect to x we get

$\displaystyle sec^2(y)\cdot \frac{dy}{dx}=1 \iff \frac{dy}{dx}=\frac{1}{\sec^2(y)}$

$\displaystyle \frac{dy}{dx}=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^ 2(y)}=\frac{1}{1+x^2}$

The last step comes from the eqaution at the top $\displaystyle \tan(y)=x$

it was really help mates, i now clearly understand how it comes