A trough is 3 meters long and its ends have the shape of isosceles triangles tha

Question

A trough is 3 meters long and its ends have the shape of isosceles triangles that are 1 meter across at the top and have a height of 0.5 meters. If the trough is being ?lled with water at a rate of 0.3 m^3 /min, how fast is the water level rising when the water is 0.25 meters deep?

Explanation / Answer

a) the quantities that are given are : Length of the trough = 10 m across at the top ( base of isosceles triangle ) = 1 m height of isosceles triangle or ( trough ) = 50 cm = 0.5 m dv/dt = 0.5m^3/min height when WATER = 0.2 m ------- b) variables are: V ---> volumes b ----> base of triangle h -----> height of triangle L --> length of the trough dv/dt ----> water rate in of the trough unknown : dh/dt -----> the water height with respect tot time -------- c) V = L * Area V = L * ( (1/2) * b * h ) i will substitute the b with 2h ( explanation as the following: ) the following picture of the isosceles triangle , we can use proportions : . . .. . _______________ . - - - - -- - - - --. I . - - - - - - - - . I . . - - - - - - . . I h = 0.5 = (1/2) . . . - - - - - . . . I . . . . - - . . . . I . . . .. - . . . . . I (b/2) / h = (1/2) / (1/2) (b/2) / h = 1 b / (2h) = 1 b = (2h) yup V = L * ( (1/2) * 2h * h ) V = L * h^2 ------. Let's start differentiating with L as constant because L is not moving -------- d) dv/dt = L * 2 * h * dh/dt ( dv/dt ) / ( 2 * L * h ) = dh/dt ------> when the height is 0.2 m ( 0.5 ) / ( 2 * 10 * 0.2 ) = dh/dt 0.125 m/min = dh/dt

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