Thread: Determining equation from given graph

1. Determining equation from given 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

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?

2. Standard Form Of a Quadratic Equation

You know the standard equation of a quadratic equation?

a(x-h)^2+k where (h,k) is the vertex of the parabola.

Well this is obviously a Quadratic because the graph is a downward parabola which is symetrical with respect to the y-axis.

Can you find a way to write this equation in standard form without anymore of my help?

Hint: h is the last term of your eqution!

3. Oh I haven't done that for so long. It is a(x)^2+h?

4. remember a is not defined in the given graph.

Spoiler:
try $\frac{-h}{b^{2}}x^{2}+h=y$
.

lol. sorry this dosent work. how do you add a spoiler?

5. Oh ok that makes more sense. I knew b^2 had to be part of it.

6. Originally Posted by ment2byours

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

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?
a.) Well the two zeroes are b and -b, and the parabola is concave down, so the equation is $f(x)=-k(x-b)(x+b)$. Since you know that $f(0)=-k(-b)(b) = kb^2 = h$, we can conclude that $k=\frac{h}{b^2}$ So the equation you're looking for is $f(x)=-\frac{h}{b^2}(x-b)(x+b)$.

b.) If it's a cosine curve, and it takes 2b to complete a half-cycle, then it takes 4b to complete a full cycle. Thus we have that the period is $\frac{2\pi}{4b}=\frac{\pi}{2b}$. So the equation is $g(x)=k\cos\left(\frac{\pi}{2b}x\right)$. Since $g(0)=k=h$, we have that $g(x)=h\cos\left(\frac{\pi}{2b}x\right)$

c.) To see which has the greatest area, simply calculate the area for each one.

$-\frac{h}{b^2}\int_{-b}^b x^2-b^2\,dx = -\frac{h}{b^2}\left[\frac{x^3}{3}-b^2 x\right]_{-b}^b = \frac{4}{3}hb$

$h\int_{-b}^b \cos\left(\frac{\pi}{2b}x\right)\,dx = h\left[\frac{2b}{\pi}\sin\left(\frac{\pi}{2b}x\right)\rig ht]_{-b}^b = \frac{4}{\pi}hb$

Comparing $\frac{4}{3}hb$ and $\frac{4}{\pi}hb$, we see that $\frac{4}{3}hb > \frac{4}{\pi}hb$, so the area of the parabola is bigger.

adhyeta: You add a spoiler by wrapping spoiler tags around it "(spoiler)" and "(/spoiler)" (using square brackets of course, but I couldn't write them properly here or otherwise the system would actually put a spoiler in this explanation!)

7. Originally Posted by redsoxfan325

adhyeta: You add a spoiler by wrapping spoiler tags around it "(spoiler)" and "(/spoiler)" (using square brackets of course, but I couldn't write them properly here or otherwise the system would actually put a spoiler in this explanation!)
i tried that. dosent work. is the code case sensitive?

8. I don't think so. In your post above, you have the close tag as [ /spoiler]. Delete the extra space to make it [/spoiler].

9. Originally Posted by redsoxfan325
I don't think so. In your post above, you have the close tag as [ /spoiler]. Delete the extra space to make it [/spoiler].
-wow-

thnx!