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Math Help - Finding an equation that is tangent to a curve and whose slope is a max.

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    Finding an equation that is tangent to a curve and whose slope is a max.

    a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

    b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

    For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
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    Quote Originally Posted by kmjt View Post
    a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

    b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

    For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
    First derivative test.
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    Differentiation once gives the slope of the function .

    So differentiation after differentiation , you would find the extrema of the slope !

    Differentiation thrice you would test whether it is max .
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    Hmm could I possibly have an example? I've never done the first derivative test with a sinusoidal function.. only basic stuff like f(x)=x^3-6x^2+15 for example
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    Quote Originally Posted by kmjt View Post
    Hmm could I possibly have an example? I've never done the first derivative test with a sinusoidal function.. only basic stuff like f(x)=x^3-6x^2+15 for example
    Suppose f(x)=sinx

    f'(x)=cos(x)=0

    Where does cos(x)=0?

    \frac{\pi}{2}+\pi k

    Thus your critical values are at those points.

    Then you check the left and right of the values to determine if it is a relative max or min.

    Since this function is wave, all the rel. max will be the same.
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    Quote Originally Posted by kmjt View Post
    For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
    Simple. Just take the derivative of that again (i.e. the second derivative). So the original function is

    y = 2cos(3x)

    then..

    \frac{dy}{dx} = -6cos(3x)

    to find when the derivative is at a maximum, we take the derivative once more

    \frac{d^2y}{dx^2} = -18sin(3x)

    and we find that it's at a maximum when x = \frac{\pi}{6}.

    So than we evaluate the first derivative to see what the slope is at \frac{\pi}{6}.

    \frac{dy(\frac{\pi}{6})}{dx} = -6

    so the slope, which happens to be the optimal slope, is -6.

    To find where on the graph we should draw the tangent line, we evaluate \frac{\pi}{6} using the regular function and we find that

    y(\frac{\pi}{6}) = 0

    using this info we can do some algebra to get the equation of this line which is

    y = -6x + \pi

    b) This isn't the only possible solution. Remember when we took the second derivative

    \frac{d^2y}{dx^2} = -18sin(3x)

    and set it equal to 0 and got x = \frac{\pi}{6}. Then we evaluated \frac{\pi}{6} using the original function, y = 2cos(3x), and we got y(\frac{\pi}{6}) = 0.

    Essentially what this is telling us is that the original function hits a slope of \pm6 when ever it touches the x-axis (equal to 0).

    Since it's a trig function, it continues to hit 0 to infinity.

    The real answer (when the function hits a maximum slope) is x  = \frac{\pi}{6} + k\frac{\pi}{3} where k is an integer.

    Cheers!
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    Quote Originally Posted by eddie2042 View Post

    to find when the derivative is at a maximum, we take the derivative once more

    \frac{d^2y}{dx^2} = -18sin(3x)

    and we find that it's at a maximum when x = \frac{\pi}{6}.
    How do you know the max is at x = \frac{\pi}{6} from looking at -18sin(3x)?
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    Junior Member eddie2042's Avatar
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    Quote Originally Posted by kmjt View Post
    How do you know the max is at x = \frac{\pi}{6} from looking at -18sin(3x)?

    uh-oh.. i just notcied a mistake. in my original post i said the derivative is

    \frac{dy}{dx} = -6cos(3x) <<<< WRONG!

    it's supposed to be


    \frac{dy}{dx} = -6sin(3x) <<<<< CORRECT!

    this means that the second derivative is also wrong. it was

    \frac{d^2y}{dx^2} = -18sin(3x) <<<< WRONG!


    it's supposed to be

    \frac{d^2y}{dx^2} = -18cos(3x)<< CORRECT!

    ok now we're back on track. We set the second derivative equal to zero, and we find the slope is at a max when

    -18cos(3x) = 0

    3x = arcos(0)

    x = \frac{\pi}{6}

    Everything's correct from there.. i mention the second derivative again; replace it with the corrected one...
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    <br /> <br />
3x = arcos(0)<br />

    I'm not quite sure what that means. I understand everything else.
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    Junior Member eddie2042's Avatar
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    Quote Originally Posted by kmjt View Post
    <br /> <br />
3x = arcos(0)<br />

    I'm not quite sure what that means. I understand everything else.
    I'm trying to solve for x. We started with the equation

     -18cos(3x)  = 0

    then we divide both sides by -18

    cos(3x)  = 0

    then we take the inverse cos function of both sides (sometimes it's written with a power of minus one but i prefer arcos. so..

     3x = arccos(0)
     3x = \frac{\pi}{2} .. inverse cos of 0 is 90 degrees (or pi/6 radians)

    and finally...

    x = \frac{\pi}{6}
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    Quote Originally Posted by kmjt View Post
    a) Find an equation of a line that is tangent to the curve f(x)=2cos3x and whose slope is a maximum.

    b) Is this the only possible solution? Explain. If there are other possible solutions, how many solutions are there?

    For a.. I know how to find the tangent of a curve at x=something.. but this question asks to find the tangent whose slope is a maximum; what would I have to do to determine that?
    this is a situation where your knowledge of trig graphs goes a long way in determining the maximum slope with little calculus manipulation.

    y = 2\cos(3x) has a period of T = \frac{2\pi}{3}, and given the general shape of the cosine curve, the maximum slope will occur when 3x = \frac{3\pi}{2} ... x = \frac{\pi}{2}.

    this maximum slope will occur at all values of x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z} , and will be equal to 6.
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    <br /> <br />
x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z}<br />

    What does that mean? I just sketched the graph of 2cos3x and I see how there is a max at every 2pi/3 interval.
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    Quote Originally Posted by kmjt View Post
    <br /> <br />
x = \frac{\pi}{2} + k \cdot \frac{2\pi}{3} \, ; \, k \in \mathbb{Z}<br />

    What does that mean? I just sketched the graph of 2cos3x and I see how there is a max at every 2pi/3 interval.
    It means that at all those points of x slope is max, for a k (constant integer) within the region Z
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    Quote Originally Posted by AllanCuz View Post
    It means that at all those points of x slope is max, for a k (constant integer) within the region Z
    region Z ?

    k \in \mathbb{Z} means k is an element of the set of integers ... \mathbb{Z} is the symbol for the set of integers.
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  15. #15
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by skeeter View Post
    region Z ?

    k \in \mathbb{Z} means k is an element of the set of integers ... \mathbb{Z} is the symbol for the set of integers.
    Yeah, my bad on that one..good thing I'm not a prof yikes
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