You are walking along a sidewalk toward a 40 foot wide sign which is adjacent to the sidewalk and

perpendicular to it (Fig. 22).

(a) If your viewing angle is 10 degrees, then how far are you from the

nearest corner of the sign?

(b) If your viewing angle is 10degrees and you are walking at 25 feet

per minute, then how fast is your viewing angle changing?

© If your viewing angle is 10 degrees and is increasing at 2degrees

per minute, then how fast are you walking?

# Please explain the answer to part b and c of this calculus question

**Hndalama**#1

**trishc**#2

I think this is right, but not 100% sure. This is an implicit differentiation problem.

(a) Since this is a right triangle, and you know one leg angle and one leg length and want to know the other leg length, use tangent.

tan theta = 40 / x, so x = 40 / tan theta

theta = 10 degrees, so x = 40/(tan 10) = 226.9 feet

(b) Both x and theta change as a function of time. Since you are walking towards the sign, x is getting smaller so

dy/dx =x’ is negative. x’ = -25 ft/min

tan theta(t) = 40/x(t)

Take the derivative with respect to time of both sides, then substitute in theta = 10, x = 40/tan theta, and x’ = -25

(secant squared theta) ( theta’ ) = (-40) (x^-2) ( x’ )

After some simplification:

theta ’ = 25 sin 10 cos 10 = 4.3 degrees / min

© Plug theta ’ = 2 degrees / min, x = 40/(tan 10), theta = 10 into the same equation as (b) and then solve for x’

(secant squared theta) ( theta’ ) = (-40) (x^-2) ( x’ )

After some simplification:

x’ = -2/[(cos 10)(sin 10)] = -11.7 ft/min