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For a National Board Exam Review:

An arc 18m high has the form of a parabola with the axis vertical. If the width of the arc 8m from the top is 64m, Find the width of the arc at the bottom.

Answer is 96

Construct Equation:

$${ (y-k) = -4a(x-h)^2 }$$

Assume ${(h,k) = (0,18) }$ and ${(x,y) = (64/2,10)}$

$${ (10-18) = -4a(32)^2 ; a = \frac{-1}{152} }$$

$${ a = \frac{1}{152} }$$

${ Change (x,y) to (x,0) }$$

$${ (0-18) = -4(\frac{1}{152})(x-0)^2 ; x=26.15 ... }$$

What am I doing wrong? Any Hint?

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  • $\begingroup$ where you wrote 18-10 should be 10-18 $\endgroup$ – Chester Aug 20 '15 at 0:54
  • $\begingroup$ @Chester edited! $\endgroup$ – james Aug 20 '15 at 1:03
  • $\begingroup$ ... which would only swap a sign, but your $a$ is incorrect as well. FYI: it is quicker to do the arithmetic for this problem if you write things in powers of 2, i.e. $8=2^3$, $4=2^2$, $32 = 2^5$. $\endgroup$ – Chester Aug 20 '15 at 1:05
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Let $w > 0$ be the desired width. Place the origin at the vertex so that the parabola has the form $y = ax^2$. Then the parabola goes through the points $(32, -8)$ and $(w/2, -18)$. Plugging in the first point yields: $$ -8 = a(32)^2 \iff a = \frac{-1}{4 \cdot 32} $$ Plugging in the second point yields: $$ -18 = \frac{-1}{4 \cdot 32} \cdot \frac{w^2}{4} \iff w^2 = (4 \cdot 4)(2 \cdot 9)(2 \cdot 16) \iff w = 4 \cdot 2 \cdot 3 \cdot 4 = 96 $$

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Supposing $a=1$, your horizontal movement when you are $8$ metres below the arc's maximum should be its square root, $2 \sqrt2$. However, the width of one half the arc is $32$, meaning that the arc is $\frac {32} {2 \sqrt2} $, or $8 \sqrt2$ times wider than if $a=1$. Similarly, at $18$ metres below the arc's maximum, the width of one half the arc should be its square root, or $3 \sqrt2$. However, we know the arc is $8 \sqrt2$ times wider than this, so we multiply the two.

$$8 \sqrt2 * 3 \sqrt2 = 24 * 2 = 48. $$

Remember, this width is only one half of the arc at its base. Thus,

$$48 * 2 = 96$$

Hope this helped.

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