Precalc question

Question

A street light is 10 feet above a straight bike path. Olav is bicycling down the path at a rate of 15 MPH. At midnight, Olav is 33 feet from the point on the bike path directly below the street light. The relationship between the intensity C of light (in candlepower) and the distance d (in feet) from the light source is given by C = k/d^2 , where k is a constant depending on the light source.

(a) From 20 feet away, the street light has an intensity of 1 candle. What is k?

(b) Find a function which gives the intensity of the light shining on Olav as a function of time, in seconds.

(c) When will the light on Olav have maximum intensity?

(d) When will the intensity of the light be 2 candles?

Answer 1

Hi Flora H.,

For a) we have d = 20 and C = 1 and equation C = k/d². Plug and chug.

1 = k/20²

k = 20² = 400 ft² *candlepower

For b) we have a speed of 15 mph, a distance of 33 ft from under the light, and 10 ft from the light to the street.

-Olav's distance from the light is √(10² + 33²). A right triangle formed by the distance from the light to the street as a vertical leg and a horizontal leg equal to Olav's distance from being under the light.

-Olav's speed in ft/s is 15 mi/hr *(5280 ft/1 mi)*(1 hr/3600s) = 22 ft/s

-Olav's speed (v) equals distance (d) over time (t), v = d/t or d = v*t = 22*t.

-Olav's distance (d_UL UL for Under Light) from being under the light is d_ul = 33 - 22*t, this is the horizontal leg of the right triangle with time (t).

-Olav's distance (d) from the light itself is d = √(10² + [33 - 22*t]²), which we can plug into the equation.

C = 400/d² = 400/{√(10² + [33 - 22*t])}² = 400/(100 +[33 - 22*t]²) candlepower, this is the function with time.

(If you forget the 10 ft for the height of the light the equation blows up when 33 - 22t = 0, which means you are in the light).

For c) Max intensity is when 33 - 22*t = 0.

33 = 22*t

t = 1.5 s

For d) Set C equal to 2 and solve for t.

2 = 400/(100 + [33 - 22*t]²)

t = 43\22 s and t = 23/22 s, once before being under the light and once after.

I hope this helps, Joe.

Answer 2

For a) given that C = k/d² we can multiple both sides by d² to get k = Cd2 then plug in the given numbers for C (which is 1) and d (which is 20) to get k = (1)(20²) or k = 400 candlepower ft²

For b) first change the units of 15 MPH (miles/hour) into feet/sec like this:

15 miles/hour x (5280 feet)/(1 mile) x (1 hour/3600 seconds) = 88/3 feet/sec. Now, using distance = speed x time, we can calculate the distance Olav is from the light as (33 - 88/3t). We can now plug that in wherever there is a "d" in the equation C = k/d² while also replacing the k with the 400 we calculated in the first question to get:

C = 400/(33 - 88/3t)² which is a perfectly fine answer but we can perhaps simplify it as follows:

Get a common denominator and combine 33 with 88/3 t:

C = 400/(99/3 - 88/3t)²

C = 400/[(99-88t)/3]² then factor out 11 and square the 11 and 3

C = 400/[(121/9)(9 - 8t)²]

C = 3600/[121(9 - 8t)²]

C = 29.752/(9 - 8t)²

then write it in function notation:

C(t) = 29.752/(9 - 8t)²

For c) since the equation of intensity is undefined at d = 0, mathematically, it doesn't really work to say it's a maximum, but logically, the highest intensity is when d = 0. Since we found that d = 33 - 88/3t, we can make d = 0 and solve for t.

0 = 33 - 88/3t

88/3t = 33

t = 33(3/88) = 9/8 = 1.125 seconds

For d) to find the time where the light intensity = 2 candlepower, make C = 2 and solve:

C(t) = 29.752/(9 - 8t)²

2 = 29.752/(9 - 8t)²

2((9 - 8t)² = 29.752

(9 - 8t)² = 29.752/2

(9 - 8t)² = 14.876

√(9 - 8t)² = ±√14.876

9 - 8t = ±3.857

solution #1: 9 - 8t = 3.857 or t = 0.643 seconds

solution #2: 9 - 8t = -3.857 or t = 1.607 seconds

(he will have this light intensity as he gets close to the light and then again on the other side of it.