## Rain Gutter

# A long sheet of tin 12 inches in width is to be bent to form a rain gutter as shown below. (You have a piece of posterboard of this length!) Determine angle theta so that the gutter will carry the most water, that is, so its cross-sectional area is at a maximum.

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# Let's consider a more general case, where the width of the tin is 3b. We transform the trapezoidal cross section above into a rectangle by cutting off the left-most triangle and putting it on the right:

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# We see that the area of the rectangle is base*height. With a little trigonometry, we see that base*height is (b+b*cosθ)*b*sinθ=b^{2}(1+cosθ)*sinθ=b^{2}(sinθ+cosθ*sinθ)=f(θ).

# Task 1. Let's try some values of θ (which we'll now call X) and see what happens. Set MODE to degree. X can range from 0° to 180°. But can you see that if X > 90°, the area gets smaller?

Try this:

From the home screen, enter

"B^{2}(sin(*L*X)+cos(*L*X)*sin(*L*X))"→A

{5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95}→X

"ΔList(*L*A)/ΔList(*L*X)"→R

R is a list of rates of change, namely, the difference of two adjacent areas divided by the difference of two adjacent angles.

Now, in STAT Edit, enter:

Name=X

Name=A

Name=R

The three lists will automatically fill. What do you see? How does the area A vary? How is R computed?

Just as with the unit on the open box, you can change the values of X and again go to STAT Edit.

# Task 2.

We want the area, f(θ), to be as big as possible. So we take the derivative and set it equal to zero. (First we may take out b^{2}, a constant. Or we can leave it in. If we leave it in, we need to assign a value to b on the home screen. We are using a 12-inch wide piece of poster board, so our b is 12/3 = 4 inches. So we may store the value 4 in B: 4→B.)

d/dt(sinθ + cosθ*sinθ) = 0

Press MODE and set the calculator to Degree.

Here is one way to proceed:

In the Y= menu, set

\Y_{1}=B^{2}(sin(X)+cos(X)sin(X))

Now use Solver:

EQUATION SOLVER

eqn:0=nDeriv(Y_{1},X,X) ENTER

B=4

X= (something; here you enter a guess)

bound={-1E99,1E99}

Since X cannot be negative, you may set the bound as follows:

bound={0,1E99}

Make a guess (for example, 45 degrees) for X, and leave the cursor on the X= row.

Press ALPHA SOLVE, and you will see

X=59.999999971637

So the trapezoid has a maximum area when the angle is 60°. But what is the area of the trapezoid when the angle is 60°? On the home screen, enter

Y_{1}

You will see

20.78460969

Of course, the units are square inches.

# For those who know some trigonometric identities, here is another way:

The derivative of sinθ = cosθ.

We use a trigonometric identity: cosθ*sinθ=½sin(2θ).

d/dt(½sin(2θ))=cos(2θ).

So d/dt(sinθ+cosθsinθ) = cosθ + cos(2θ) = 0

We know that cosθ + cos(180 - θ) = 0.

Therefore cos2θ = cos(180 - θ)

So now let's ask when 180-θ = 2θ.

3θ = 180°, so θ = 60°.

# Task 3. you may also investigate when the area of the trapezoid is greatest by setting the window as follows:

Xmin=0

Xmax=180

Xscl=10

Ymin=0

Ymax=30

Yscl=5

Xres=1

Then graph Y_{1}.

# Use CALC to locate the maximum.

# So again we see the maximum area is when the angle is sixty degrees.

Here is the rain gutter cross section, drawn to scale:

#

# Can you make your posterboard into a "gutter" with angles to maximize the water flow?

Webpage Maintained by Owen Ramsey

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