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calculus--please helpp

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if the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k, find k.

all i know is that its related to finding a derivative and i m basically completely lost....can please someone provide me some help?????

  • calculus--please helpp - ,

    the derivative gives the slope of the tangent line to the curve, at any point (x,y)

    The line 3x-4y=0 has slope= 3/4

    At any point on the curve (x,x^3 + k) the slope of the tangent line is 3x^2

    So, you want to find x where 3x^2 = 3/4

    x = 1/2

    so, now you have y = 1/2^3 + k = 1/8 + k

    What point on the line has y = 1/8 + k?

    3* 1/2 - 4(1/8 + k) = 0
    3/2 - 1/2 - 4k = 0
    4k = 1
    k = 1/4

    So, y = x^3 + 1/4

    At x = 1/2, y = 1/8 + 1/4 = 3/8

    Note that at the point (1/2 , 3/8) 3x-4y = 0

  • calculus--please helpp - ,

    Note on missing answer.
    x can also be -1/2 if 3x^2 = 3/4

    In that case, k = -1/4
    and the point of tangency is (-1/2 , -3/8)


    Note also, that if the line had been 3x-4y=1, the same line would have touched the curve at both points, with k=0.

  • calculus--please helpp - ,

    thanks so much!

  • calculus--please helpp - ,

    1/4

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