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Math Help - Find the slope of a line tangent to a curve.

  1. #1
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    Find the slope of a line tangent to a curve.

    I encountered this problem months ago in a math competition, and have yet to have any understanding how it could be solved.

    A non-vertical line with y-intercept 2 is tangent to the curve given by the equation x^2 + y^2 = 12x + 20y - 100. Find the slope of the line.

    This problem would be easy if it was single variable, but because it has both x and y, I have zero clue where to start.
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    Newbie Phantasma's Avatar
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    Re: Find the slope of a line tangent to a curve.

    x^2+y^2=12x+20y-100. Let's use the chain rule: \frac{d}{dx}\left[x^2+y^2\right]=2x+2y\frac{dy}{dx}=\frac{d}{dx}\left[12x+20y-100\right]=12+20\frac{dy}{dx}. Solving for \frac{dy}{dx}, we get \frac{dy}{dx}=\frac{12-2x}{2y-20}=\frac{6-x}{y-10}. Can you take it from there?

    Edit: My apologies. I didn't see that this was in Pre-Calculus. If you haven't learned derivatives yet, you'll need some trigonometry.
    Last edited by Phantasma; July 15th 2013 at 12:08 PM.
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    Re: Find the slope of a line tangent to a curve.

    Start by figuring out the circle's characteristics. Then you can finish with a bit of trigonometry.
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    Re: Find the slope of a line tangent to a curve.

    In other words, it is an "implicit function"- y is defined "implicitly" by that formula- you could, theoretically, solve for y.

    There are two ways to continue. One, the simpler in concept but harder in practice, is to actually solve for y as a function of x: x^2+ y^2= 12x + 20y -100 can be written as y^2- 20y+ x^2- 12x+ 100= 0 and now treat that as a quadratic equation in y with x just some number: by the quadratic formula, y= \frac{-20\pm\sqrt{400- 4(x^2- 12x+ 100}}{2}= \frac{-20\pm\sqrt{48x- 4x^2}}{2}= -10\pm\sqrt{12- x^2} and differentiate that.

    The other way, more "sophisticated" but easier in practice, is to use "implicit differentiation". By the chain rule the derivative of y^2 with respect to x is the derivative of y^2 with respect to y times the derivative of y with respect to x: 2yy'. And the derivative of 20y with respect to x is 20y'. Differentiating both sides of x^2+ y^2= 12x+ 20y- 100 with respect to x gives 2x+ 2yy'= 12+ 20y'. Solving for y', 20y'- 2yy'= (20-2y)y'= 2x- 12, y'= \frac{2x- 12}{20- 2y}= \frac{x- 6}{10- y}.

    In either case, you still have the problem that you do not know at what "(x,y)" you want to have the tangent line. Instead, replace x with some letter, say, a, and y with its value in terms of that "a". Write the equation of the tangent plane and set the value of y, when x= 0, equal to 2. The solve that equation for a.
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    MHF Contributor MarkFL's Avatar
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    Re: Find the slope of a line tangent to a curve.

    I would let the tangent line be:

    y=mx+2

    With the given curve, complete the square on the two variables to obtain:

    (x-6)^2+(y-10)^2=6^2

    Now, substitute for y into the equation, then require the discriminant to be zero (do you understand why?). Now solve for m.
    Last edited by MarkFL; July 15th 2013 at 12:25 PM.
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    Re: Find the slope of a line tangent to a curve.

    I do know derivatives, but I put it in precalculus because I knew they would not be complex for this problem. I got (6-x) / (y-10) as the slope, however, that is not enough information to find the slope as we have no point to give, and therefore cannot create a slope when the only info on the tangent line I have is that it takes the form y = mx + 2. I'm still exploring the thought of trigonometry.
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    Newbie Phantasma's Avatar
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    Re: Find the slope of a line tangent to a curve.

    Quote Originally Posted by ShadowKnight8702 View Post
    I do know derivatives, but I put it in precalculus because I knew they would not be complex for this problem. I got (6-x) / (y-10) as the slope, however, that is not enough information to find the slope as we have no point to give, and therefore cannot create a slope when the only info on the tangent line I have is that it takes the form y = mx + 2. I'm still exploring the thought of trigonometry.
    Your BIG hint for today is to consider what m is. You'll have two equations and two unknown variables.
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    Re: Find the slope of a line tangent to a curve.

    Ah, I figured it out. you have that m = dy/dx and you have m = (y-2)/x. Thanks.
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