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Math Help - Find the equation of the tangent line to the graph

  1. #1
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    Find the equation of the tangent line to the graph

    Hi everyone. I'm working on a calc problem, but I found it to be difficult. Maybe I'm missing the point.

    Find the equation of the tangent line to the graph of f(x)=3x tan^2(x) + x^2 cos(x)

    Well, this is the derivative I got:

    3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)

    Since it just says the equation of the line, but it doesn't specify an x-value, so I just plug in a random value. However, I found it extremely difficult to do it the regular way, because trig functions are involved.

    Is there a trick to solve this problem?

    Thanks a lot.
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  2. #2
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    m=3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)

    Where m is the slope of the tangent line to the curve.

    The key word is equation.
    Last edited by dwsmith; January 16th 2011 at 08:00 PM. Reason: forgot m
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  3. #3
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    You can leave it as \displaystyle x=x_0 for a future value, then \displaystyle  y-f(x_0)=f'(x_0)(x-x_0)
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    I would just write

    Let m=3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)

    and as pickslides said let (x_0,y_0) \in f(x)

    y-y_0=m(x-x_0)\Rightarrow y=m(x-x_0)+y_0

    This way you can save yourself the trouble of writing out m.
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  5. #5
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    What value should I choose for x?
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    Quote Originally Posted by zhangxupage2 View Post
    What value should I choose for x?
    The arbitrary value x_0
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    If I made x zero, then the slope is going to be zero also. The equation would become a horizontal line.
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    Quote Originally Posted by zhangxupage2 View Post
    If I made x zero, then the slope is going to be zero also. The equation would become a horizontal line.
    No one said to make it zero. That is the slope at x = 0. Does your question ask for the slope at x = 0?
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    What should I make x of? I'm not sure what you meant by arbitrary x0?
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    Quote Originally Posted by zhangxupage2 View Post
    What should I make x of? I'm not sure what you meant by arbitrary x0?
    Call it z than.
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  11. #11
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    Maybe I haven't learned it in your way. Can I please explain it in a simpler way? Thanks a lot.
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  12. #12
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    Quote Originally Posted by zhangxupage2 View Post
    Maybe I haven't learned it in your way. Can I please explain it in a simpler way? Thanks a lot.
    I am sorry. I don't know how else to explain it.
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  13. #13
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    Thanks a lot. I'm trying to find a value for x, but I don't know which one to use.
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  14. #14
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    Quote Originally Posted by zhangxupage2 View Post
    Hi everyone. I'm working on a calc problem, but I found it to be difficult. Maybe I'm missing the point.

    Find the equation of the tangent line to the graph of f(x)=3x tan^2(x) + x^2 cos(x)

    Well, this is the derivative I got:

    3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)

    Since it just says the equation of the line, but it doesn't specify an x-value, so I just plug in a random value. However, I found it extremely difficult to do it the regular way, because trig functions are involved.

    Is there a trick to solve this problem?

    Thanks a lot.
    As it has been said already, you need to evaluate your derivative at an arbitrary point to get the slope of your line.

    So far, you've correctly shown that f^{\prime}(x)=3\tan^2x+6\tan x\sec^2 x+2x\cos x-x^2\sin x.

    Now, by convention, we let x=x_0 be the arbitrary point (for clarification x_0 and 0 are not the same thing!) and substitute this into your derivative to see that

    m=f^{\prime}(x_0)=3\tan^2x_0+6\tan x_0\sec^2 x_0+2x_0\cos x_0-x_0^2\sin x_0

    Now this value for the slope will change for any x_0; in other words, x_0 is a place holder for a specific x value when its defined in the problem. If the problem had said that you needed to find the equation of the tangent line at x=3, x=2\pi, or x=\frac{1}{2}\pi, then you would replace x_0 with 3, 2\pi or \frac{1}{2}\pi respectively.

    So now, in general, your tangent line at the arbitrary value x=x_0 will have the form y-f(x_0)=m(x-x_0). Thus the answer should be:

    y=\left(3\tan^2x_0+6\tan x_0\sec^2 x_0+2x_0\cos x_0-x_0^2\sin x_0\right)(x-x_0)+3x_0\tan^2x_0+x_0^2\cos x_0

    Certainly, this equation can be simplified when an actual value is given. Since one wasn't given, we have to leave it in this form.

    I hope this clarifies things.
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  15. #15
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    Quote Originally Posted by zhangxupage2 View Post
    What should I make x of? I'm not sure what you meant by arbitrary x0?
    This values acts as a temporary value for what you might want to make \displaystyle x .

    In your case you were not given an x-value to use in the question. That is O.K. Nothing to worry about.

    I think the confusion here is the fact you have a \displaystyle x and an \displaystyle x_0 in the following equation.

    \displaystyle y-f(x_0)=f'(x_0)(x-x_0)

    The \displaystyle x_0 is were you will substitute in a co-ordinate where the x is just part of the equation of the straight line. No need to replace him.

    I understand this is a little confusing.

    For your own sake pick a value i.e. \displaystyle x=1 and finish question like normal. What do you get? Think about what it means.

    It might help you bring the bigger picture together.
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