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Math Help - tangent problem

  1. #1
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    tangent problem

    An interesting question.

    Find the equation of the tangent for the curve  y = x^2 at the point  (1,1) without using calculus.
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    Quote Originally Posted by danny arrigo View Post
    An interesting question.

    Find the equation of the tangent for the curve  y = x^2 at the point  (1,1) without using calculus.
    you mean without using differentiation, right? well, if so, suppose the line is y=mx + b. since (1,1) belongs to the line, we have b=1-m. now the tangent line should intersect the curve

    at one point only, i.e. the equation x^2 - mx + m-1=0 must have only one solution. so the discriminant has to be 0, which gives us m=2. thus the line is y=2x-1.
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    Okay, so how about  y = x^n say for some positive integer  n ?
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    Quote Originally Posted by danny arrigo View Post
    Okay, so how about  y = x^n say for some positive integer  n ?
    since x = 1 must be the root of the equation x^n-mx+m-1=0 of multiplicity at least 2, we'll have: x^n - mx + m - 1=(x-1)^2p(x), for some polynomial p(x). thus:

    (x-1)(x^{n-1}+ \cdots + x + 1 - m)=(x-1)^2p(x), and hence: x^{n-1} + \cdots + x + 1 - m = (x-1)p(x). not put x=1 to get m=n, and therefore b=1-n. so the equation

    of the tangent line is y=nx + 1-n.
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     y = \sin x or  y = e^x at say  (0,0) or  (0,1), resp.?
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    Quote Originally Posted by danny arrigo View Post
     y = \sin x or  y = e^x at say  (0,0) or  (0,1), resp.?
    Why do you think they invented Calculus!
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  7. #7
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    Quote Originally Posted by danny arrigo View Post
     y = \sin x or  y = e^x at say  (0,0) or  (0,1), resp.?
    1. \sin(x) at (0,0). We know the tangent line is \hat{y}-0=\cos(0)(x-0)\implies \hat{y}=x

    Ok so taking a cue from NonCommAlg and assuming that you mean no derivatives this is simple. It can be easily show using non caluculus terms that the function \frac{\sin(x)}{x} approaches one as x approaches zero. So now consider this let f be a differentiable function and \hat{y} its tangent line at a point x=c. Consider that a tangent line approximates a curve in a neighborhood of c, and that \hat{y}(c)=f(c). So it is obvious that \frac{f(x)}{\hat{y}} must get closer and closer to one as x gets closer and closer to c. So we must have that \frac{\sin(x)}{mx} must get closer and closer to one as x gets closer and closer to zero.But because as we noted \frac{\sin(x)}{x}\to 1 we must have that \frac{\sin(x)}{mx}\to 1 \implies m=1 so the tangent is y=x.


    I implictly used calculus I suppose.

    I think for the second one I can do better

    For the function f(x)=e^xwe know that the tangent line, \hat{y}, at (0,1) must be of the form \hat{y}-1=mx\Rightarrow \hat{y}=mx+1. Now consider the tangent line \hat{y}_2 of the function f_1(x)=e^{-x} at (0,1). It is clear geometrically that these functions have perpendicular slopes, so the tangent line must be of the form \hat{y}_2=\frac{-x}{m}+1. But now consider that f_1(-x)=f(x) in this case so \hat{y}(x)=mx+1=\frac{x}{m}+1=\hat{y}_2(-x)\implies m=1

    Im not too sure about that one...can someone verify?
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