Question

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 )

Answer 1

man you are just in love with these huh?

Answer 2

looks like
\[\frac{12}{x}=\tan(\theta)\] it the relation you can use

Answer 3

take the derivative, get
\[\frac{-12}{x^2}x'=\sec^2(\theta)\theta'\]

Answer 4

you know
\[x'=-5\] and here it is -5 because x is decreasing

Answer 5

replace x by 9, and you should be good to go

Answer 6

haha not in love at all thats why i need you!!

Answer 7

where am i suppose to replace x

Answer 8

x is 9

Answer 9

hard part is finding
\[\sec(\theta)\] when x is 9, but not that hard