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Math Help - help plz

  1. #1
    flewoveru
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    help plz

    Find an equation for the line tangent to the curve at the given point
    x^4+y^3=17 at (-3,-4)

    Also if you could post the steps it took you to get the answer plz THANKS GUYS!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Differentiate implicity the function, then plug (x,y)=(-3,-4), so y' will be the slope of the tangent line.

    Apply the point-slope formula and you're done.

    --

    I dunno if this is the procedure, I don't remember these things -.-
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by flewoveru View Post
    Find an equation for the line tangent to the curve at the given point
    x^4+y^3=17 at (-3,-4)

    Also if you could post the steps it took you to get the answer plz THANKS GUYS!
    As Krizalid said:
    x^4+y^3=17

    4x^3 + 3y^2 \frac{dy}{dx} = 0

    \frac{dy}{dx} = - \frac{4x^3}{3y^2}

    So at (-3, -4) the derivative is:
    \frac{dy}{dx} = - \frac{4(-3)^3}{3(-4)^2} = \frac{4 \cdot 27}{3 \cdot 16} = \frac{9}{4}

    Now you know the tangent line at (-3, -4) has a slope of 9/4. Thus
    y = mx + b

    -4 = \frac{9}{4} (-3) + b

    b = -4 + \frac{27}{4} = \frac{11}{4}

    So your line is
    y = \frac{9}{4}x + \frac{11}{4}

    -Dan
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