anonymous
  • anonymous
A person 6 ft tall is watching a streetlight 18 ft high while walking toward it at a speed of 5 ft/sec (see figure 3.53 in the book). At what rate is the angle of elevation of the person's line of sight changing with respect to time when the person is 9ft from the base of the light? (Let x(t) denote the distance from the lamp at time and the angle of elevation at time t )
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
man you are just in love with these huh?
anonymous
  • anonymous
|dw:1332705328926:dw|
anonymous
  • anonymous
looks like \[\frac{12}{x}=\tan(\theta)\] it the relation you can use

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More answers

anonymous
  • anonymous
take the derivative, get \[\frac{-12}{x^2}x'=\sec^2(\theta)\theta'\]
anonymous
  • anonymous
you know \[x'=-5\] and here it is -5 because x is decreasing
anonymous
  • anonymous
replace x by 9, and you should be good to go
anonymous
  • anonymous
haha not in love at all thats why i need you!!
anonymous
  • anonymous
where am i suppose to replace x
anonymous
  • anonymous
x is 9
anonymous
  • anonymous
hard part is finding \[\sec(\theta)\] when x is 9, but not that hard
anonymous
  • anonymous
|dw:1332705886528:dw|
anonymous
  • anonymous
you need the hypotenuse to find \[\sec(\theta)\] and you have a triangle with 9 and 12 so hypotenuse is 15, and therefore \[\sec(\theta)=\frac{15}{9}\]
anonymous
  • anonymous
plug those in to this one \[-\frac{12}{x^2}x'=\sec^2(\theta)\theta'\] and solve for \[\theta'\]
anonymous
  • anonymous
man i did that and i got -2.06 but thats not right

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