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Math Help - triangles and rates of change (and some other stuff)

  1. #1
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    triangles and rates of change (and some other stuff)

    1)A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?

    Okay, so here's what I did

    dx/dt=5 x=30

    y=16

    need to find ds/dt (shadow)

    s/6=(s+x)/16

    16s=6s+6s

    s=(3/5)x thus (3/5)=ds/dt

    That's what my teacher told me to do, but I have no idea why I did that and I STILL don't know how to get the answers. I understand that I probably have to make a smaller triangle where the y is 6 ft, but I just don't know where to go from there.

    2)Linear approximation

    Use linear approximation, i.e. the tangent line, to approximate as follows: cubed root 63.6
    The equation of the tangent line to at f(x) x=64 can be written in the form y=
    I don't even understand what the equation is, but shouldn't the answer be close to 4?

    I'll probably have some more questions as I work through this homework, but I really think that I'll start to understand it if someone helps me with these few problems. Thanks guys!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nobodygirl View Post
    1)A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?

    Okay, so here's what I did

    dx/dt=5 x=30

    y=16

    need to find ds/dt (shadow)

    s/6=(s+x)/16

    16s=6s+6s

    s=(3/5)x thus (3/5)=ds/dt

    That's what my teacher told me to do, but I have no idea why I did that and I STILL don't know how to get the answers. I understand that I probably have to make a smaller triangle where the y is 6 ft, but I just don't know where to go from there.
    See the diagram below, ALWAYS DRAW A DIAGRAM WHEN DOING A RELATED RATES PROBLEM.

    you were right about forming a smaller triangle within the larger one with a height of 6 ft. where do you go from there? well, you use the concept of similar triangles. remember similar triangles are triangles whose sides are propotional, so the ration of any two sides of one triangle is equal to the ratio of the corresponding sides of a triangle similar to it.

    so the height of the big triangle is the 16ft pole with the light on top of it. the 6 ft line x meters away from the pole is the woman (she's on the Atkin's diet which is why she's so thin).

    Obviously, we will use similar triangles here.

    Now 16/(x + y) = 6/y ……the ratio of height/base of the large triangle = ratio of height/base of small triangle
    => 16y = 6x + 6y
    => 10y = 6x …………….now differentiate implicitly
    => 10 dy/dt = 6 dx/dt = 6(4) = 24
    => dy/dt = 24/10 = 2.4 ft/sec

    But the rate at which the tip of her shadow is moving away from the pole is equal to the rate at which she is moving plus the rate at which her shadow is extending in front of her. So then,

    ds/dt = dx/dt + dy/dt = 4 + 2.4 = 6.4 ft/sec


    so this is a slightly different way from how your professor did the problem, well not really, it is essentially the same, except he solved for y where i just left 10y.

    i'm kind of in a rush now, i'll do the second problem later
    Attached Thumbnails Attached Thumbnails triangles and rates of change (and some other stuff)-tri.gif  
    Last edited by Jhevon; March 28th 2007 at 07:13 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nobodygirl View Post

    2)Linear approximation

    Use linear approximation, i.e. the tangent line, to approximate as follows: cubed root 63.6
    The equation of the tangent line to at f(x) x=64 can be written in the form y=
    I don't even understand what the equation is, but shouldn't the answer be close to 4?
    Ok, so linear approximation is not that hard, all you have to do is memorize one formula. the formula is:

    f(x) ~= f(a) + f ' (a)(x - a)

    x is the value that you want to find f(x) for but obviously can't, while a is a value close to x that you do know the value of f(a) for and you will use to approximate f(x). let's see how it works.

    you want to find (63.6)^(1/3), (this is the same as the cube root of 63.6, i just wrote it in terms of powers). so we don't know what (63.6)^(1/3) is, however, 64 is very close to 63.6 and we do know what (64)^(1/3) is, that's just 4. so we want x = 63.6 and a = 64, now let's do the problem.

    Consider the function f(x) = x^(1/3), we want to approximate f(63.6), that is, (63.6)^(1/3).

    now f ' (x) = (1/3)x^(-2/3)

    choose x = 63.6 and a = 64

    => f(a) = 64^(1/3) = 4
    => f ' (a) = (1/3)64^(-2/3) = (1/3)(1/64^(2/3)) = (1/3)(1/16) = 1/48
    => (x - a) = 63.6 - 64 = -0.4

    so using f(x)~= f(a) + f '(a)(x - a), we have

    (63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(4) = 47/12 = 3.9167

    the actual value is 3.991649245, pretty darn close if you ask me
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    Quote Originally Posted by Jhevon View Post
    ...

    so using f(x)~= f(a) + f '(a)(x - a), we have

    (63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(4) = 47/12 = 3.9167

    the actual value is 3.991649245, pretty darn close if you ask me
    Hello, Jhevon,

    I don't want to pick at you but you have made a small but nevertheless effective mistake:

    (63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(0.4) = 479/120 = 3.99167

    And now your last remark makes more sense

    EB
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earboth View Post
    Hello, Jhevon,

    I don't want to pick at you but you have made a small but nevertheless effective mistake:

    (63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(0.4) = 479/120 = 3.99167

    And now your last remark makes more sense

    EB
    ah, thanks for that, i had 4 instead of 0.4

    yet another proof someone like me shouldn't be doing math 1am in the morning

    my mistake wasn't really far off from the answer though, wait! i shouldn't be smiling damn stupid mistakes!

    sorry nobodygirl


    P.S. I think its great how we look out for each other on this site!!
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    Thanks you guys, this really helped a lot!
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