The great Pyramid of Egypt has a square base of 230 meters. To estimate the height h of this massive structure, an observer stands at the midpoint of one pf the sides and views the apex of the pyramid. The angle of elevation alpha is found to be 52 degrees. How accurate must this measurement be to keep the error in h between -1 meter and 1.
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The Chain rule
- Thread starter serenity
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serenity said:The great Pyramid of Egypt has a square base of 230 meters. To estimate the height h of this massive structure, an observer stands at the midpoint of one pf the sides and views the apex of the pyramid. The angle of elevation alpha is found to be 52 degrees. How accurate must this measurement be to keep the error in h between -1 meter and 1.
First write the height as a function of the angle of elevation.
H = 115 * Tan(x)
Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
BigGlenntheHeavy
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\(\displaystyle Given: \ h \ = \ 115tan(\theta), \ \theta \ = \ 52^{0} \ = \ \frac{13\pi}{45}r, \ dh \ = \ \pm \ 1 \ meter\)
\(\displaystyle Hence, \ dh \ = \ 115sec^{2}(\theta)d\theta, \ \pm1 \ = \ 115sec^{2}\bigg(\frac{13\pi}{45}\bigg)d\theta, \ d\theta \ = \ \pm.0033r\)
\(\displaystyle Ergo, \ \theta \ must \ be \ kept \ within \ \frac{13\pi}{45}\pm.0033 \ radians \ in \ order \ for \ the \ error \ in \ height \ to \ be \ within \ \pm1 \ m.\)
\(\displaystyle Check: \ h \ = \ 115tan(13\pi/45) \ = \ 147.19 \ meters\)
\(\displaystyle h \ = \ 115tan(13\pi/45+.0033) \ = \ 148.19 \ meters\)
\(\displaystyle h \ = \ 115tan(13\pi/45-.0033) \ = \ 146.19 \ meters\)
\(\displaystyle Hence, \ dh \ = \ 115sec^{2}(\theta)d\theta, \ \pm1 \ = \ 115sec^{2}\bigg(\frac{13\pi}{45}\bigg)d\theta, \ d\theta \ = \ \pm.0033r\)
\(\displaystyle Ergo, \ \theta \ must \ be \ kept \ within \ \frac{13\pi}{45}\pm.0033 \ radians \ in \ order \ for \ the \ error \ in \ height \ to \ be \ within \ \pm1 \ m.\)
\(\displaystyle Check: \ h \ = \ 115tan(13\pi/45) \ = \ 147.19 \ meters\)
\(\displaystyle h \ = \ 115tan(13\pi/45+.0033) \ = \ 148.19 \ meters\)
\(\displaystyle h \ = \ 115tan(13\pi/45-.0033) \ = \ 146.19 \ meters\)