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Math Help - Calculus

  1. #1
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    Calculus

    Hi i am having trouble with three questions in my calculus assignment, all help is greatly appreciated

    Find the equation of the tangent to y= x+lnx which is perpendicular to the line x+2y+3=0. I am not really sure where to start with this one :X

    A piece of wire one meter long is to be cut into two pieces one will be used to form a square the other a circle, find the dimensions of the two that will form the least total area.
    I'm not sure if this is correct but i subed x=25-.5piR into x^2+piR^2, which gave me (25-.5piR)^2 + piR^2

    Determine the stationary points of the curve y=(x^2-1)e^-x. Using the product rule i come up with y'= -(x^2 -1)e^-x+2xe^-x , i am unsure whether i am right up to this point or if it can be simplified any further.

    Thanks,
    Webby
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    Quote Originally Posted by Webby View Post
    Hi i am having trouble with three questions in my calculus assignment, all help is greatly appreciated

    Find the equation of the tangent to y= x+lnx which is perpendicular to the line x+2y+3=0. I am not really sure where to start with this one :X
    f(x)=x+ln(x).
    f '(x)=1+1/x

    The equation of the tangent to the curve, at a point of absciss a is y=f '(a)(x-a)+f(a)
    The only unknown is a.
    Since this tangent is perpendicular to x+2y+3=0, the slopes of these two lines have to be inverse to each other (i.e. n and 1/n). The slope of the tangent is f '(a).

    A piece of wire one meter long is to be cut into two pieces one will be used to form a square the other a circle, find the dimensions of the two that will form the least total area.
    I'm not sure if this is correct but i subed x=25-.5piR into x^2+piR^2, which gave me (25-.5piR)^2 + piR^2
    See here and here for similar problems.

    Determine the stationary points of the curve y=(x^2-1)e^-x. Using the product rule i come up with y'= -(x^2 -1)e^-x+2xe^-x , i am unsure whether i am right up to this point or if it can be simplified any further.
    Your derivative is correct. Now, factorise as much as possible
    There is an obvious common factor in y'.
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  3. #3
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    Quote Originally Posted by Webby View Post
    Find the equation of the tangent to y= x+lnx which is perpendicular to the line x+2y+3=0.
    The second equation can be rearranged to become y=-0.5x-1.5 so the slope of the line is 0.5 and you are looking for a tangent that has a slope of 1/0.5 or 2.

    Now any tangent has a slope given by the differential of y. So differentiate y=x+lnx to get y'=1+1/x

    This must equal 2 so set it equal to 2 and solve for x (answer x = 1).

    Put x=1 into y=x+lnx to find y. Now you know the slope of the tangent line and you know one point through which it must pass. From this information you can work out its equation.

    Quote Originally Posted by Webby View Post
    A piece of wire one meter long is to be cut into two pieces one will be used to form a square the other a circle, find the dimensions of the two that will form the least total area.
    I'm not sure if this is correct but i subed x=25-.5piR into x^2+piR^2, which gave me (25-.5piR)^2 + piR^2
    Let the amount of wire allocated to the square be x. Therefore, 1-x is allocated to the circle.

    Each side of the square is x/4 long so the area of the square is x^2/16.

    The circumfrence of the circle is 1-x so it has a radius of (1-x)/pi/2 and its area is pi[(1-x)/pi/2]^2=(1-x)^2/pi/4

    The total area A = x^2/16+(1-x)^2/pi/4

    Now we have an equation for the area which we can minimise by:
    1) differentiate it with respect to x
    2) set the differential equal to zero
    3) solve for x.
    4) One of the two solutions will give the minimum area.

    Quote Originally Posted by Webby View Post
    Determine the stationary points of the curve y=(x^2-1)e^-x. Using the product rule i come up with y'= -(x^2 -1)e^-x+2xe^-x
    Looks good so far. Now set equal to zero and solve for x.

    (x^2-1)e^{-x}=2xe^{-x}
    (x^2-1)=2x
    (x^2-2x-1)=0
    Solve this for x to get the value for x at the two stationary points. Then put the values of x into the original equation to find the value of y at the stationary points.
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