# Related rates word problem

#### bkbowser

##### New member
I am assuming that I am setting this up incorrectly or substituting the incorrect values.

A water trough is 10 m long and a cross-section has the shape
of an isosceles trapezoid that is 30 cm wide at the bottom,
80 cm wide at the top, and has height 50 cm. If the trough is
being filled with water at the rate of 0.2 m^3͞/min, how fast is the
water level rising when the water is 30 cm deep?

I broke up the trapezoid into 2 triangles and a rectangle so the area should be

A = 2*Triangle + Rectangle
A = (b*h)+(50*h)

I can look at b in terms of h by using similar triangles. Each triangle has side a, of 25 side b, of 50. So the triangle formed by the water level must be side a 15 side b of 30. So b is h/2

A = (h^2*1/2) + (50*h)

Converting this into volume should be easy, just multiply the equation A, by 1000.

dA/dt = 2/2*h*dh/dt + 50*dh/dt

Convert dA/dt into dV/dt;

1000*dA/dt = dh/dt(h+50)*1000

Since 1000*dA/dt I can make a substitution

dV/dt = dh/dt(h+50)*1000

I have a book answer of 10/3. Could someone be so kind as to point out the errors? Thanks!

#### awkward

##### Member
It might be a good idea to double-check your formula for the area of the trapezoid in terms of h. It should yield the area of the whole trapezoid when h = 50. Does it?

• bkbowser

#### bkbowser

##### New member
(80+50)/2*50=3250

versus

25*50+30*50=2750

so no they are not the same. i can't think of any particular reason why they are not equal though. and there's the added difficulty of it gumming the whole rest of the problem up #### HallsofIvy

##### Well-known member
MHB Math Helper
I am assuming that I am setting this up incorrectly or substituting the incorrect values.

A water trough is 10 m long and a cross-section has the shape
of an isosceles trapezoid that is 30 cm wide at the bottom,
80 cm wide at the top, and has height 50 cm. If the trough is
being filled with water at the rate of 0.2 m^3͞/min, how fast is the
water level rising when the water is 30 cm deep?

I broke up the trapezoid into 2 triangles and a rectangle so the area should be

A = 2*Triangle + Rectangle
A = (b*h)+(50*h)
You are told the the height of the trapezoid is 50 cm so why are you multiplying h by 50? The base of the trapezoid is 30 cm wide so the area of the central square is 30h, not 50h.

I can look at b in terms of h by using similar triangles. Each triangle has side a, of 25 side b, of 50. So the triangle formed by the water level must be side a 15 side b of 30. So b is h/2

A = (h^2*1/2) + (50*h)

Converting this into volume should be easy, just multiply the equation A, by 1000.

dA/dt = 2/2*h*dh/dt + 50*dh/dt

Convert dA/dt into dV/dt;

1000*dA/dt = dh/dt(h+50)*1000

Since 1000*dA/dt I can make a substitution

dV/dt = dh/dt(h+50)*1000

I have a book answer of 10/3. Could someone be so kind as to point out the errors? Thanks!

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