Related Rates Calculus 1

Question

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 11 ft3/min, how fast is the water level rising when the water is 8 inches deep?

Answer 1

Let V = volume of water in the trough at time t

 

Let h = depth of water at time t

 

Let x = length of the surface of the water at the triangular end of the trough at time t

 

Given:  dV/dt = 11     Find:  dh/dt when h = 8 in = 2/3 ft

 

Since the triangular end is isosceles, the altitude bisects the base.

 

Area of triangular cross section of the filled portion of the trough = (1/2)(x)(h)

 

   By similar triangles (comparing the triangular end of the trough to the triangle formed by the filled portion), we have 2/1 = (x/2)/h.  So, x = 4h. 

 

   So, Area of triangular cross section = (1/2)(x)(h) = (1/2)(4h)(h) = 2h²

 

V = (Area of triangular cross section)(length of trough) =  (2h²)(10) = 20h²

 

dV/dt = 40h(dh/dt) 

 

Substituting the given information, we have 11 = 40(2/3)(dh/dt)

 

                                                                11 = (80/3)(dh/dt)

 

                                                               dh/dt = 33/80 ft/min

 

Answer 2

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