A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Calculus 1: The illumination of an object by a light source is directly proportional to the strength of the sourcee and inversely proportional to the square of the distance from the source. if 2 light sources, one 3 times as strong as the other, are placed 10 ft apart, where should the object be placed on the line between the sources so as to receive the least illumination
anonymous
 5 years ago
Calculus 1: The illumination of an object by a light source is directly proportional to the strength of the sourcee and inversely proportional to the square of the distance from the source. if 2 light sources, one 3 times as strong as the other, are placed 10 ft apart, where should the object be placed on the line between the sources so as to receive the least illumination

This Question is Closed

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.01. Choose a coordinate system on that line between the two sources. 2. Write down the light intensity as a function in these coordinates. You can take all proportionality constants as 1 because they don't influence extremal points. 3. Find the local and then global extrema of that function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do I write down the light intensity as a function?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You have to write the distance in terms of your coordinate and add the intensities of the two lamps together.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.1to get you started we know our function needs to be in terms of distance because thats what they want as an answer. define variables I = illumination s=strength of lamp 1 d=distance from lamp 1 direct proportion means s goes up, I goes up inverse proportion means d^2 goes up, I goes down Lamp 1: I = s/d^2 Lamp 2: I = 3s/(10d)^2 second lamp 3X as strong and total space between lamps is 10 so distance from lamp 2 is 10d Total Illumination: like nowhereman said we have to add them together I = s/d^2 + 3s/(10d)^2 Now the goal is to minimize illumination Differentiate with respect to d and set equal to 0 solve for d
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.