Question on rates of change

Question

A water trough is 7 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 60 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 30 cm deep?

Answer 1

Hi Alex, 

 

    Let's start by finding a formula for the volume of a trough (drawing a picture always helps): In general,  

Volume = length × width × height, where length = 7 meters in this case and width × height is the area of the end piece (aka isosceles trapezoid).  

 

    The formula for the area of the end piece is Area = (a + b)/2 × h , where a is the length of the top part, b is the length of the bottom part and h is the height.  In this case, b = 20 cm.  But as the trough fills one can think of "a" in the water-filled portion as changing from 20 cm to 60 cm as "h" changes from 0 to 40 cm.  (In the jargon of calculus  one could say . . . wait for it . . . wait for it . . .   a = f(h). )    

 

   a = f(h) is a linear function where a = 20 cm + (60 - 20)/(40 - 0) × h (cm) or simply a = 20 + h  (in cm)

 

SHORT ANSWER:  At the instant in question, surface area of the water in the trough is 7m×0.5m=3.5m²  and the height is increasing at a rate dh/dt.  Thus dV/dt = (Area)×dh/dt.  Putting in dV/dt = 0.3 m³/min and the surface area, one has dh/dt = 0.086 m/min or 8.6 cm/min

 

 LONG ANSWER:  Whew!  Almost done.  Now that we have done all this stuff,  what was the original question ?  Oh,  given dV/dt = 0.3 m³/min,  what is dh/dt when h = 30 cm ? 

 

V = l × (a+b)/2 × h = 7 x (0.20  + h + .0.20)/2 x h    = 7 x (h²/2 + 0.20h)    (units of m³)

 

dV/dt = 7(h + 0.2)dh/dt   now put in dV/dt = 0.3, h = 0.30 and solve for dh/dt = 0.086 m/min or 8.6 cm/min

 

WOW!  THE SAME ANSWER!  ISN'T PHYSICS WONDERFUL ?

Answer 2

Hi Alex,

 

This question can be solved using related rates analysis, a powerful tool that any calculus student should be able to use and understand.  In this problem, we are given a physical object (a pool of water) whose geometry is changing in some known way (the volume of water in the pool is increasing at a certain rate).  We write what we know about the pool in this way:

dV/dt = 0.3 m³/min

 

To relate this to the geometry of the pool, we need to solve for the volume of the water in the trough.  We will assume that the trough is level, meaning that water spreads evenly across the 7m by 20cm base of the trough.  To solve for the volume of the trough when the water level is a certain height, we will multiply the length of the trough by the cross-sectional area.

 

First, we need to find the cross-sectional area as a function of the height of the water.  Splitting the trough into a central rectangular portion (where the base of the trough is flat) and two right triangular portions (where the wall of the trough is slanted) gives the following breakdown for the area:

A(h) = A_rectangle(h) + 2*A_triangle(h)

 = 20cm * h + 2*(1/2)*(base_triangle)*h

 

The part the triangle is a little bit harder to fill in, because we don't know base_triange, the width of the triangles.  In fact, since the trough is slanted, this width will change as a function of the height.  The wall is just a linear slope, though, and we know the proportions of the resulting triangle, so we can compute this width by setting up a proportionality between similar right triangles:

base_triangle / h = [(60cm - 20cm)/2] / 40cm

   = 20cm/40cm

∴ base_triangle = h * 20cm/40cm = h/2

 

Now, we can write the volume of the water in the trough when the water level is h cm deep as the length of the trough multiplied by the cross-sectional area:

V(h) = length * A(h)

 = 7m * [A_rectangle(h) + 2 * A_triangle(h)]

 = 7m * [20cm * h + 1/2 * h²]

 

Here, we recognize that the height is changing in time, and we can relate the rate of change of the water volume to the changing height by computing the derivative of V(h) implicitly with respect to t:

dV/dt = 7m * [20cm * (dh/dt) + h * (dh/dt)]

 

Finally, we plug in what we know about dV/dt and h from the problem:

0.3 m³/min = 7m * [20cm * (dh/dt) + (30cm) * (dh/dt)]

 

This is a linear equation of one variable which can be solved for dh/dt to give the final answer to the problem:

dh/dt = (0.3 m³/min)/[ 7m * (20cm + 30cm) ]

  = (0.3 m³/min)/[ 3.5 m² ]

  = 0.0857 m/min

 

In this problem, we used the geometry of the problem along with calculus to set up a relationship between known rates of change and unknown quantities.  Using similar principles, you can reason about interesting problems involving trigonometric relations, physics of motion, and many, many other geometry problems of related rates!