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Math Help - Tangent question :)

  1. #1
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    Talking Tangent question :)

    Hey, anyone good at tangents?

    10(iv) Given that 24x+3y+2=0 is the equation of the tangent to the curve at the point (p,q), find p and q. (The curve is y=1/3x-9x.)

    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by LoveDeathCab View Post
    I'm not sure, i have done some calculus. Whats the other option?
    To me the other option is vague, and i hit a wall. i found two possible x-values for p, but i can't really tell (or don't remember how to tell) which is the right one without calculus.

    Here's how i found the possible x-values, maybe you or someone else can figure out which is the right one without calculus.


    p and q are just the x and y values (respectively) for which the line and the curve intersect.

    now 24x + 3y + 2 = 0
    => 3y = -24x - 2
    => y = -8x - 2/3

    now equate the functions:
    since y = -8x - 2/3 and y = 1/3x-9x
    then they intersect when

    -8x - 2/3 = (1/
    3)x - 9x
    =>(1/3)x^3 - x + 2/3 = 0
    => x^3 - 3x + 2 = 0
    we see that x = 1 is a root, so x - 1 is a factor. divide by x - 1. we obtain:
    x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)(x + 2)(x - 1) = 0

    so x = 1 or x = -2
    these are the possible values for p, at one of these points we have the tangent, i just can't find a simple way to figure out which i don't want to use calculus since you're doing AS math and you're not comfortable with it

    i can tell you the answer is x = p = 1, and you could just plug that value into either of the other functions to get q, but i used calculus to come to that conclusion
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  3. #3
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    Hang on a sec! I remember, calculus is differenciation and intergration. I am comfortable with that, all that dy/dx etc! Is there a way for getting the right answer using calculus?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by LoveDeathCab View Post
    Hang on a sec! I remember, calculus is differenciation and intergration. I am comfortable with that, all that dy/dx etc! Is there a way for getting the right answer using calculus?
    really? is calculus required for the math you are doing? anyway, i will show you how i came up with the answer using calculus, but i think there is a way to know without it.

    now we found the equation of the line to be:
    y = -8x - 2/3
    this is a line with slope -8

    now all we need to do, is find the points on the cubic graph where the slope is -8. whichever value corresponds with the values we got before is the right one. (Remember, we said the tangent is either at x = 1 or x = -2)

    let's find the formula for the slope of the curve at any value of x. we use the derivative for this. that is, dy/dx gives the slope of the tangent line for any value of x.
    y = (1/3)x^3 - 9x ..............take the derivative, we get:
    => dy/dx = x^2 - 9

    we want x^2 - 9 = -8 ........that is, we want the slope of a point on the cubic to be equal to the slope of the tangent line. so,

    x^2 - 9 = -8
    => x^2 - 1 = 0
    => (x + 1)(x - 1) = 0
    => x = -1 or x = 1

    now the line only cuts the curve at x = 1 and x = -2. since it does not cut the curve at -1, x = 1 must be our answer.

    so x = p = 1

    now to find q

    we have
    y = -8x - 2/3, so when x = 1 we have,
    y = -8 - 2/3 = -26/3

    so (p,q) = (1, -26/3)

    Hope you could follow that. if not, wait on one of the smart people to give you an easier way
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  5. #5
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    How come x=-2 is not the answer? Just out of interest cause it doesn't say why? Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by LoveDeathCab View Post
    How come x=-2 is not the answer? Just out of interest cause it doesn't say why? Thanks.
    okay. so for the first part of the question, all i did was to find where the line and the curve intersect. that is, at x = 1 and x = -2. only one of these is the point that the tangent touches the curve. for the other point, the line actually cuts the curve (you can see this by using a graphing program to graph both functions, a good one is Graph )

    when i did the second part, i found what points on the curve had the same slope of the line. x = -2 was not such a point, so i discarded it. for the other point, x = -1, the line we were talking about did not even touch the curve (since the line ONLY touches the curve at x = -2 and x = 1 as we found), so that could not be it either. so the only possible answer is x = 1
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