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Math Help - Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

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    Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    Pi is confusing me deeply and I really need help... A short explaination about how to use Pi would be greatly appreciated!
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    for the tangent you need a line of the form y = mx+c

    m is the gradient of tan(x) at the point, pi/4.. this can be found by differentiating tan(x) with respect to x

    then to find c, you have a point on the line so substitute into your equation and you can find c.

    The gradient of the normal is -1/m, where m is the gradient of teh tangent
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    The number \pi is confusing you? It is about 3.141592... but is irrational and its decimal expansion continues infinitely. Historically, the first use of \pi was as the ratio of a circle to a diameter. From that we can eventually get things like sin(\pi)= 0, sin(\pi/4)= \sqrt{2}/2, cos(\pi/4)= \sqrt{2}/2, tan(\pi/4)= 1.

    I was a bit surprised, on opening a thread titled "Finding the Tangent and normal line", to find a question about the number \pi. Generally one learns about trig functions and their values, especially at integer fractions of \pi well before a Calculus class.

    The derivative of tan(x) is sec^2(x)=\frac{1}{cos^2(x)}. So the derivative, and slope of the tangent line, of tan(x) at x= \pi/4 is sec^2(\pi/4)= \frac{1}{cos^2(\pi/4)}= \left(\frac{2}{\sqrt{2}}\right)^2= 2. The tangent line to tan(x) at x= \pi/4 has slope 2 and passes through the point (\pi/4, tan(\pi/4)= (\pi/4, 1). The normal line, perpendicular to the tangent line, has slope -1/2 and passes through the point (\pi/4, 1).
    Last edited by HallsofIvy; April 2nd 2014 at 12:49 PM.
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    i have been looking at this for a while, and i can't figure why the slope is equal to 2...
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    By definition, the derivative of a function at a point is the slope of the tangent line at that point. HallsofIvy just showed you how to obtain the derivative, then how to plug in the point to find the value of the slope of the tangent line at that point. What is troubling you?
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    I might be doing this wrong, but the answer of (2/sqr(2))^2 is equal to 1. I know the answer is 2, but I can't figure to get it right
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    Here is some algebra you should know: $\left( \dfrac{a}{b} \right)^c = \dfrac{a^c}{b^c}$

    Applying that: $\left( \dfrac{2}{\sqrt{2}} \right)^2 = \dfrac{2^2}{(\sqrt{2})^2} = \dfrac{4}{2} = 2 \neq 1$
    Thanks from marco213
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    Re: Finding the Tangent and normal line of tan(x) at point (pi/4 , f(pi/4) )

    I might be missing something, but why pi/4 became 2/sqr(2) , as it was mentionned pi/4 = sqr(2)/2?
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