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Math Help - analysis vector

  1. #1
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    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?
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  2. #2
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    Hello, noob!

    Have your forgotten your differentiation formulas?


    Could anyone please tell me how to differentiate this equation?

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

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

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

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

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, noob!

    Have your forgotten your differentiation formulas?



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

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

    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
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by noob View Post
    im just confusing on how does the g(t) is obtain, maybe got some formula did i forgot
    y=tan^{-1}(x) \iff \tan(y)=x

    Now taking the implict derivative with respect to x we get

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

    \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 \tan(y)=x
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post

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

    Now taking the implict derivative with respect to x we get

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

    \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 \tan(y)=x

    i was really help mates, i now clearly understand how it comes
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post

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

    Now taking the implict derivative with respect to x we get

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

    \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 \tan(y)=x

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