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Math Help - basic calculus problems

  1. #1
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    basic calculus problems

    my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him... i was hoping that someone here could help. his homework is do tomorrow. so, time is essential.. well, i was hoping that someone could hopefully show me how to do these problems so i could show my son how to do them.

    Thanks in advance to all that this may concern.

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  2. #2
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    Quote Originally Posted by philipsk View Post
    my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him...
    Does he know that you are getting help or does he think that you know?

    1) f(x)=x-\sin 2x
    f'(x)=(x)'-(\sin 2x)'=1-2\cos 2x (chain rule).

    5) y=\frac{\sin x}{4+\cos x}
    Quotient rule,
    y'=\frac{(\sin x)'(4+\cos x)-\sin x(4+\cos x)'}{(4+\cos x)^2}
    y'=\frac{\cos x(4+\cos x)-\sin x(-\sin x)}{(4+\cos x)^2}
    y'=\frac{4\cos x+\cos^2 x+\sin^2 x}{(4+\cos x)^2}=\frac{4\cos x+1}{(4+\cos x)^2}
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Does he know that you are getting help or does he think that you know?
    philipsk is a dad he is entitled to our help to appear omnicient in the eyes
    of his chidren.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    philipsk is a dad he is entitled to our help to appear omnicient in the eyes
    of his chidren.
    I should have added that he should not say that he is getting help to his son, I forgot to.
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  5. #5
    Grand Panjandrum
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    2. Find the equation of the tangent to the curve y=\frac{\sqrt{x}}{x+6} at (4,0.2)

    The equation of a line (and the tangent is a line) is of the form y=mx+c, and this is a tangent at the given point on the curve if m is equal to the slope or gradient of the curve at the point and c is chosen so that the line passes through the point.

    Now the slope of the curve at a point is equal to the derivative of y with respect to [tex][x/math] at the point. To evaluate this derivative we need the quotient rule:

    \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2},

    with f(x)=\sqrt{x} and g(x)=x+6, so f'(x)=\frac{1}{2 \sqrt{x}}, and g'(x)=1

    Hence:

    \frac{dy}{dx}=\frac{\frac{x+6}{2\sqrt{x}}-\sqrt{x}}{(x+6)^2}

    Now we could simplify this but I think it better to just evaluate it as it is at x=4, which I make: \left. \frac{dy}{dx}\right|_{x=4}=0.005.

    So our line has the form: y=0.005\,x+c. Now when x=4\ y=0.2, so: 0.2=0.005\,4+c, hence c=0.18, so finally we have the equation of the tangent at the given point is:

    y=0.005\,x+0.18

    RonL

    (now I expect one of the other helpers will point out some dreadful mistake in my algebra )
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  6. #6
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    2. Find the equation of the line tangent to y = \frac{\sqrt{x}}{x+6} at the point (4, 0.2).

    The equation of any line may be expressed as
    y = mx + b
    where m is the slope of the line and b is the y-intercept (0, b).

    The slope of the line tangent to y = \frac{\sqrt{x}}{x+6} at the point (x, y) is the first derivative of y with respect to x. So by the quotient rule:
    y' = \frac{\frac{1}{2} \cdot \frac{1}{\sqrt{x}} \cdot (x + 6) - \sqrt{x} \cdot (1)}{(x+6)^2} <-- Mulitply top and bottom by 2\sqrt{x}

    y' = \frac{x + 6 - 2x}{2 \sqrt{x} (x + 6)^2}

    y' = \frac{-x + 6}{2 \sqrt{x} (x + 6)^2}

    So the slope of the tangent line to the curve at the point (4, 0.2) is:
    m = \frac{-4 + 6}{2 \sqrt{4} (4 + 6)^2} = 0.005 <-- Or 1/200.

    So the equation of the tangent line at (4, 0.2) is:
    y = 0.005x + b

    Now we need to find b, so put in the point coordinates:
    0.2 = 0.005 \cdot 4 + b

    b = 0.18 <-- Or 9/50

    Thus the equation of the line is:
    y = \frac{1}{200}x + \frac{9}{50}

    I have attached a graph of the function and the tangent line.

    -Dan

    Quote Originally Posted by CaptainBlack View Post
    (now I expect one of the other helpers will point out some dreadful mistake in my algebra )
    At least we agree!
    Attached Thumbnails Attached Thumbnails basic calculus problems-tangent.jpg  
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  7. #7
    Forum Admin topsquark's Avatar
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    3. Find the equation of the line tangent to y = x^8cos(x) at ( \pi, -\pi^8).

    Same method as before.

    y' = 8x^7cos(x) + x^8 \cdot - sin(x)

    y' = -x^8sin(x) + 8x^7cos(x)

    At the point ( \pi, -\pi^8) this is:
    m = -(\pi)^8sin(\pi) + 8(\pi)^7cos(\pi) = 0 - 8\pi^7 = -8\pi^7

    So we have the tangent line:
    y = (-8\pi^7)x + b
    at the point ( \pi, -\pi^8). Again, we need b, so:

    -\pi^8 = (-8\pi^7) \cdot \pi + b

    b = -\pi^8 + 8\pi^8 = 7\pi^8

    Thus the tangent line is:
    y = (-8\pi^7)x + 7\pi^8

    Again, I have attached a graph. (The scale of the y-axis hides some of the features of the graph.)

    -Dan
    Attached Thumbnails Attached Thumbnails basic calculus problems-tangent2.jpg  
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  8. #8
    Eater of Worlds
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    For #7, polynomials are among the easiest functions to differentiate. They're everywhere continuous and very well-behaved.


    \frac{d}{dx}[2x^{3}-3x^{2}-36x+4]=6x^{2}-6x-36=x^{2}-x-6

    Factor:

    (x-3)(x+2)

    As you can see, you have zeros at x=3 and x=-2

    See the graph. Note the horizontal tangents as the maximum and minimum.
    Last edited by galactus; November 24th 2008 at 06:39 AM.
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