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Calculus/Area under a curve.

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Expert: - 5/2/2009


Question

Problem's graph
An arched window with base width 2b and height h is set into a wall. The arch
is to be either an arc of a parabola or a half-cycle of a cosine curve.

a. If the arch is an arc of a parabola, write an equation for the parabola
relative to the coordinate system shown in the figure. (x-intercepts are
(−b,0) and (b,0). y-intercept is (0,h).)
b. If the arch is a half-cycle of a cosine curve, write an equation for the
cosine curve relative to the coordinate system shown in the figure.
c. Of these two window designs, which has the greater area? Justify your
answer.

a.) I got y=-x^2+b^2?
b.) I know that it might be something like cos(pi*bx/2) possibly?
c.) I'm clueless how to do this one. Do i take the integral of each equation?

Answer
Questioner:   Jeanne
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Determining equation from given graph
Question:  

An arched window with base width 2b and height h is set into a wall. The arch is to be either an arc of a parabola or a half-cycle of a cosine curve.

a. If the arch is an arc of a parabola, write an equation for the parabola relative to the coordinate system shown in the figure. (x-intercepts are (−b,0) and (b,0). y-intercept is (0,h).)


b. If the arch is a half-cycle of a cosine curve, write an equation for the cosine curve relative to the coordinate system shown in the figure.


c. Of these two window designs, which has the greater area? Justify your answer.

a.) I got y=-x^2+b^2?
b.) I know that it might be something like cos(pi*bx/2) possibly?
c.) I'm clueless how to do this one. Do i take the integral of each equation?

......................
Hi, again, Jeanne,

a. your parabolic function has  x = b, x = -b as roots, so it has (x - b)(x + b) as factors, and it has the form:

y = a(x - b)(x + b)

But the y-intercept is y = h.  Put x = 0:

h = a(-b)(b)

a = -h/b^2

So the equation is:

y = (-h/b^2)(x - b)(x + b)

y = (-h/b^2)(x^2 - b^2)  or some equivalent.

..................................

b. Your cosinusoidal function (yuk!) has this form:

y = A cos(fx + p)

[Yes, there is vocabulary for A,b,p, but who cares?]

Now if the max is  y = h, that makes  A = h.

y = A cos(fx + p)

Now the intercepts should be at   +- pi/2, but they are at  +- b

f( b) + p =  pi/2

f(-b) + p = -pi/2

Aha! Two simultaneous equations, which we can solve for f and p.

Add:

2p = 0, yes, the phase angle is zero, as we suspected. (What you mean, WE, kemosabe?)

And  f = pi/2b

OK, we have our equation:

y = h cos(pi x/2b)


Check: x = b -->

y = h cos(pi/2) = 0, right.
..........................

c. Now just integrate each one from -b to b [Or from 0 to b, just as good.]


Parabola:

{b
| (-h/b^2)(x^2 - b^2) dx
}0


(-h/b^2)(x^3/3 - xb^2) from 0 to b:

(-h/b^2)[(b^3/3 - b^3) - (0)]

(-h/b^2)[- 2b^3/3]

+ h [2b/3]

= 2bh/3

...........................
Cosine:

{b
| h cos(pi x/2b) dx
}0

= h(2b) sin(pi x/2b) / pi  from 0 to b

= 2bh/pi [(sin(pi b/2b)) - (sin(pi 0/2b))]

= 2bh/pi [(sin(pi/2)) - (sin(0)]

= 2bh/pi [1 - 0]

= 2bh/pi

OK, which is bigger?  2/3 or 2/pi?  You will answer that.

Fun.

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

Experience

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

Education/Credentials
(See above.)

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