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Math Help - Calculus / Geometry questions

  1. #1
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    Calculus / Geometry questions

    Hey guys, please help me with some of these questions, I'm really clueless how to solve them. Any help is very much appreciated.

    1: A ship S leaves a Port P at 2 pm and it sails in the direction 30T at 12km/h. Another ship is located 100 km East of P and is sailing towards P at 8km/h. (t is the number of hours after 2 pm)

    a: Show that the distance D(t), between the two ships is given by D(t) = Square root of: 304t-2800t+10000

    b: Find the minimum value of [D(t)] for all t>0.

    c: At what time are both ships closest?


    2:A fence AB is 1m high and 2m away from a wall, RQ. An ladder PQ is placed on the ence and touches the ground at P and the wall at Q.

    a: If AP = x, find the value of QR in terms of x.
    b: The ladder has the length L(x), show that [L(x)] = (x+2)(1+(1/x)

    Show that derivative[L(x)] with respect to x = 0, only when x = third square root of 2.

    d: Find the shortest length of the ladder and prove that it is the shortest length.


    3: A (symmetrical) gutter is made from a sheet of metal (30 cm wide) by bending it twice. For alpha as indicated:

    a: Show that the cross-sectional area is given by A = 100cosalpha(1+cosalpha)
    b: Using the result from a, show that derivative A with respect to alpha = 0 when sin alpha = 0.5 or -1.
    c: What value has alpha when the gutter has its maximum carrying capacity?

    Here I made a sketch for 2 of the problems:

    [img=http://img229.imageshack.us/img229/520/skizzenn6.th.jpg]

    [img=http://img229.imageshack.us/img229/1664/skizze2so9.th.jpg]
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  2. #2
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    Quote Originally Posted by Instigator View Post
    Hey guys, please help me with some of these questions, I'm really clueless how to solve them. Any help is very much appreciated.

    1: A ship S leaves a Port P at 2 pm and it sails in the direction 30T at 12km/h. Another ship is located 100 km East of P and is sailing towards P at 8km/h. (t is the number of hours after 2 pm)

    a: Show that the distance D(t), between the two ships is given by D(t) = Square root of: 304t-2800t+10000

    b: Find the minimum value of [D(t)] for all t>0.

    c: At what time are both ships closest?


    ...
    hello,

    use sin(30) = 0.5 and cos(30) = 0.5*√(3).
    Then the position of the ship S is S(6t, 6 \cdot \sqrt{3} \cdot t)

    The second vessel V is located on the east axis and it's position is V(100-8t, 0)

    Now use the distance formula:

    (D(t))^2 = (100-8t-6t)^2+(6\sqrt{3} \cdot t)^2 = 10000 - 2800t + 196t^2 + 108t^2

    The minimum of (D(t)) occurs if the first derivative of (D(t)) is zero:

    ((D(t))^2)' = 608t - 2800. Now 608t - 2800 = 0~\iff~t = \frac{2800}{608} \approx 4.6~h

    Plug in this value into the equation and calculate D: D(4.6) \approx 59.6~km
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  3. #3
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    Hello, Instigator!

    Here's #3 . . .


    3) A symmetrical gutter is made from a sheet of metal (30 cm wide) by bending it twice.
    Code:
          A     E             F     B
          * - - * - - - - - - * - - *
           \    :             :    /
            \   :             :   /
          10 \  :             :  / 10
              \α:             :α/
               \:             :/
                * - - - - - - *
                D     10      C

    For \alpha as indicated:

    (a) Show that the cross-sectional area is given by: . A \:= \:100\cos\alpha(1 + \cos\alpha) . not true!
    The area of a trapezoid is: . A \:=\:\frac{h}{2}(b_1 + b_2)

    From the right triangles, we have: . DF = EC = 10\sin\alpha\;\text{ and }\;EA = FB = 10\cos\alpha

    . . h \,=\,10\cos\alpha,\;b_1 \,=\,10,\;b_2\,=\,10+20\sin\alpha

    Therefore: . A \;=\;\frac{1}{2}(10\cos\alpha)(10 + 10 + 20\sin\alpha)\quad\Rightarrow\quad\boxed{ A \;=\;100\cos\alpha(1 + \sin\alpha)}



    (b) Using the result from (a), show that derivative of A with respect to \alpha
    equals zero when \sin\alpha \:= \:0.5\text{ or }-1
    Differentiate: . A' \;=\;100\cos\alpha(\cos\alpha) + 100(-\sin\alpha)(1 + \sin\alpha) \;=\;100\left(\cos^2\!\alpha - \sin\alpha - \sin^2\!\alpha\right)

    . . =\;100\left(1-\sin^2\!\alpha - \sin\alpha - \sin^2\!\alpha\right) \;=\;100\left(1 - \sin\alpha - 2\sin^2\!\alpha\right) \;=

    Factor and equate to zero: . (1 - 2\sin\alpha)(1 + \sin\alpha) \:=\:0


    And we have two equations to solve:

    . . 1 - 2\sin\alpha \:=\:0\quad\Rightarrow\quad\boxed{\sin\alpha \:=\:\frac{1}{2}}

    . . 1 + \sin\alpha \:=\:0\quad\Rightarrow\quad\boxed{\sin\alpha \:=\:-1}



    (c) What value of \alpha produces maximum carrying capacity for the gutter?
    Since maximum A occurs when \sin\alpha = \frac{1}{2}, then: . \boxed{\alpha \:=\:\frac{\pi}{6}}

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  4. #4
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    Quote Originally Posted by Instigator View Post
    ...

    2:A fence AB is 1m high and 2m away from a wall, RQ. An ladder PQ is placed on the ence and touches the ground at P and the wall at Q.

    a: If AP = x, find the value of QR in terms of x.
    b: The ladder has the length L(x), show that [L(x)] = (x+2)(1+(1/x)

    Show that derivative[L(x)] with respect to x = 0, only when x = third square root of 2.

    d: Find the shortest length of the ladder and prove that it is the shortest length.
    ...
    hello,

    first draw a sketch of the situation (see attachment)

    to a: You are dealing with 2 similar right triangles. Set up the proportion:

    \frac1x = \frac{\overline{QR}-1}{2}~\iff~\overline{QR} = \frac2x + 1

    to b:
    Use Pythagoran theorem:

    (L(x))^2 = (\overline{QR})^2 + (x+2)^2~\implies~(L(x))^2 = (\frac2x + 1)^2 + (x+2)^2 ~\implies~ (L(x))^2 = (\frac{2+x }{x})^2 + (x+2)^2 = (2+x)^2 \cdot \left(\frac{1 }{x}\right)^2 + (x+2)^2 . Factor the last sum and you'll get:

    (L(x))^2 = (2+x)^2 \cdot \left(\left(\frac1x\right)^2+1\right)

    to c:

    to calculate the first derivative use product rule:

    ((L(x))^2)' = \left(\frac{1 }{x^2}\right) \cdot 2 \cdot (x+2) + (x+2)^2 \cdot (-2) \cdot \left(\frac{1 }{x^3}\right) = \frac{2(2+x)(x^3-2)}{x^3}

    ((L(x))^2)' = 0~\implies~x+2 = 0~\vee~x^3-2 = 0 that means:

    x = -2~\vee~x=\sqrt[3]{2}

    If x = -2 then the ladder is placed directly against the wall and it is standing upright. If you try to use it in this position you are going to fall on your back. The only possible solution is therefore x=\sqrt[3]{2}

    to d:
    Plug in the value x=\sqrt[3]{2} into the equation of (L(x)) and calculate L(\sqrt[3]{2}). I've got

    L(\sqrt[3]{2}) = \sqrt{\frac{(\sqrt[3]{2}+2)^3}{2}}\approx \sqrt{17.32~ m^2}\approx 4.16~m

    There are several possible methods to prove that you have actually got the shortest length. Since I don't know which method you are used to use I'll leave this part of the problem to you.
    Attached Thumbnails Attached Thumbnails Calculus / Geometry questions-leiter_anwand.gif  
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  5. #5
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    I got it, thank you both very much for your help! I appreciate it very much.
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