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Thread: Further Differentiation rates of changes

  1. #1
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    Further Differentiation rates of changes

    a horse trough has a triangular cross section of height 25 cm and base 30 cm, and 2m long. A horse drinking steadily and when the water level is 5 cm below the top it is being lowered at the rate of 1 cm/ min. find the rate of consumption in litres per minute


    My attempt:
    v = $\displaystyle \frac{1}{2}bh \times l $
    thus v =
    $\displaystyle b = \frac{6h}{5}$
    $\displaystyle v = \frac{6h^2}{5}$
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} h} =\frac{12h}{5} $
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= \frac{12h}{5} \times \frac{\mathrm{d}h }{\mathrm{d} t} $
    since the water level is lowered by 5 cm thus h = 20 cm
    thus;
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= 48 \times \frac{\mathrm{d}h }{\mathrm{d} t} $


    $\displaystyle \frac{\mathrm{d}c }{\mathrm{d} t}=\frac{\mathrm{d}c }{\mathrm{d} h} \times \frac{\mathrm{d}h }{\mathrm{d} t} $
    im stuck here
    please help


    thank you
    Last edited by bigmansouf; Feb 24th 2019 at 08:02 PM.
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  2. #2
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    Re: Further Differentiation rates of changes

    Quote Originally Posted by bigmansouf View Post
    a horse trough has a triangular cross section of height 25 cm and base 30 cm, and 2m long. A horse drinking steadily and when the water level is 5 cm below the top it is being lowered at the rate of 1 cm/ min. find the rate of consumption in litres per minute


    My attempt:
    v = $\displaystyle \frac{1}{2}bh \times l $
    thus v =
    $\displaystyle b = \frac{6h}{5}$
    $\displaystyle v = \frac{6h^2}{5}$ Watch your units … length = 2m=200cm … v =120h^2
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} h} =\frac{12h}{5} $
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= \frac{12h}{5} \times \frac{\mathrm{d}h }{\mathrm{d} t} $
    since the water level is lowered by 5 cm thus h = 20 cm
    thus;
    $\displaystyle \frac{\mathrm{d}v }{\mathrm{d} t}= 48 \times \frac{\mathrm{d}h }{\mathrm{d} t} $


    $\displaystyle \frac{\mathrm{d}c }{\mathrm{d} t}=\frac{\mathrm{d}c }{\mathrm{d} h} \times \frac{\mathrm{d}h }{\mathrm{d} t} $
    im stuck here
    please help


    thank you
    see red above
    Also what is c? You should define all your variables. (Recall that 1 cm^3 of water has a volume of 1mL.)
    Thanks from topsquark and bigmansouf
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  3. #3
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    Re: Further Differentiation rates of changes

    Because the sides are straight lines, for any height the base is a linear function of height: b= ch for some constant c. The base of the triangular cross-section is 30 cm when the height is 25 cm so 30= c(25)- c= 30/25= 6/5. The area of the triangular cross section is (1/2)(6h/5)(h)= (3/5)h^2 square cm. The volume of water in the trough when the water height is "h" cm is (3/5)h^2(200)= 120h^2 cubic cm. The rate of change then is the derivative with respect to time so is 240h dh/dt cubic cm per min. You are told that h= 25- 5= 20 cm and dh/dt= -1 cm/min. Putting those numbers into the 120 h^2 will give the rate at which the trough is being emptied in cubic cm per minute, 240(20)(-1)= -4800 cubic cm per minute. You will need to convert from cm per minute to liters per minute.
    Thanks from bigmansouf
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