A photographer uses a light meter to measure the intensity of light from a flashbulb. The intensity for the flashbulb, I, in lux, is a function of the distance from the light, d, in metres, and can be represented by I(d)=10/d^2,d>0
a) Determine the following, to two decimal places.
i) the intensity of light 3 m from the flash bulb
ii) the average rate of change in the intensity of light for the indicated 1<d<3
iii) the approximate instantaneous rate of change in the intensity of light at exactly 3 m from the flash bulb
b) What does the sign of your answer to part a), subpart iii), indicate about the light intensity?
Here is what I have done
ai) l(d)=10/d^2
l(3)=10/3^2
l(3)=1.11
aii) AROC = y2-y1/x2-x1
=10-1.1111/1-3
=8.8889/-2
=-4.44445
=-4.44
I just need help on aiii) and b
a)
i) To find the intensity of light 3 m from the flash bulb, we substitute d = 3 into the function I(d) = 10/d^2:
I(3) = 10/3^2
I(3) = 10/9
I(3) ≈ 1.11 lux
ii) To find the average rate of change in the intensity of light for 1 < d < 3, we need to find the difference in intensity between d = 1 and d = 3, and divide by the difference in distance:
ΔD = 3 - 1 = 2
ΔI = I(3) - I(1) = 10/9 - 10/1^2 = 10/9 - 10/1 = 10/9 - 90/9 = -80/9
Average rate of change = ΔI / ΔD = (-80/9) / 2 ≈ -4.44 lux/m
iii) To find the approximate instantaneous rate of change in the intensity of light at exactly 3 m from the flash bulb, we need to find the derivative of the function I(d) = 10/d^2, and substitute d = 3:
I'(d) = -20/d^3
I'(3) = -20/3^3
I'(3) = -20/27
I'(3) ≈ -0.74 lux/m
b) The negative sign of the answer to part a), subpart iii) indicates that the light intensity is decreasing at a rate of approximately 0.74 lux/m when at a distance of exactly 3 m from the flash bulb.
a) To determine the answers to these questions, we'll need to use the given function:
I(d) = 10/d^2, for d > 0.
i) To find the intensity of light 3 meters from the flashbulb, substitute d = 3 into the function:
I(3) = 10/(3^2) = 10/9 ≈ 1.11 lux.
Therefore, the intensity of light 3 meters from the flashbulb is approximately 1.11 lux.
ii) The average rate of change in the intensity of light for the indicated 1 < d < 3 can be found by calculating the difference in intensity and distance at the two endpoints of that interval.
Average rate of change = (change in intensity) / (change in distance)
For this interval, the endpoints are d = 1 and d = 3.
Intensity at d = 1: I(1) = 10/(1^2) = 10 lux.
Intensity at d = 3: I(3) = 10/(3^2) = 10/9 lux.
Change in intensity = I(3) - I(1) = (10/9) - 10 = -80/9 ≈ -8.89 lux.
Change in distance = 3 - 1 = 2 meters.
Average rate of change = (-8.89 lux) / (2 meters) ≈ -4.445 lux/meter.
Therefore, the average rate of change in the intensity of light for the interval 1 < d < 3 is approximately -4.445 lux/meter.
iii) To approximate the instantaneous rate of change in the intensity of light at exactly 3 meters from the flashbulb, we can find the derivative of the function with respect to d.
I'(d) = -20/d^3
Substitute d = 3 into this derivative function:
I'(3) = -20/(3^3) = -20/27 ≈ -0.741 lux/meter.
Therefore, the approximate instantaneous rate of change in the intensity of light at exactly 3 meters from the flashbulb is approximately -0.741 lux/meter.
b) The sign of the answer to part a) subpart iii) indicates that the light intensity is decreasing at a rate of approximately 0.741 lux/meter as the distance from the flashbulb increases.