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Math Help - Tangent Lines Perpendicular

  1. #1
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    Tangent Lines Perpendicular

    find the equation of the line tangent to the curve y=x^4 and perpendicular to 3x-96y=2.

    I have thought about it for a while now, and the only part i can think of that might help solve this is that the perpendicular slope is the negative reciprocal of the 3x-96y=2, which would mean the slope is -64.

    Any help is much apreciated
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  2. #2
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    Re: Tangent Lines Perpendicular

    Hey DylanRenke.

    There is a trick with two-dimensional spaces in that m1*m2 = -1 where m1 is the gradient of one line and m2 is the gradient of the perpendicular line. So if you have an equation y = m1*x + b then a perpendicular line to y will be of the form y_perp = -(1/m1)*x + b2.

    So this means the tangent must be equal to -(1/m1) where m1 is the gradient obtained by simplifying 3x-96y=2 into the appropriate form to get m1.
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  3. #3
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    Re: Tangent Lines Perpendicular

    That is exactly what i did to obtain the number -64, but im not sure how to get the 'b' value in the y=mx+b equation so that it is tangent to y=x^4
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  4. #4
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    Re: Tangent Lines Perpendicular

    Re-arranging the formula 3x-96y=2 gives us 96y = 3x - 2 which means y = (3/96)x - 2 in which y = mx + b so m1 = (1/32) and b = -2. This means the perpendicular line will have a gradient of -(1/m1) = -(1/(1/32)) = -32. so any line perpendicular to the first line will have the form y = -32x + b for some y-intercept b.
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  5. #5
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    Re: Tangent Lines Perpendicular

    I think you've calculated the slope of 3x-96y=2 wrong. If you put it in the form y=mx+b, you should get m=\frac{3}{96}=\frac{1}{32}, so then the slope of the perpendicular is -32, not -64.

    Now you need to find where the tangent to y=x^4 has slope -32. The slope of the tangent line to a curve y=f(x) is just the derivative (you're learning this in Calculus, right?). If you take the derivative and go through the algebra, you should get x=-2. So y=(-2)^4=16. So a point on the line is (-2,16), and the slope is -32.

    Now plug into the point-slope formula for a line: y-y_0=m(x-x_0), so y-16=-32(x+2).

    I left a few gaps on purpose for you to fill in. Post back if you're still having trouble.

    - Hollywood
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