Help Derivatives - maximizing sunlight through a window

In summary, the conversation discusses how the amount of daylight a location receives can be modeled by a sinusoidal function and how to calculate the maximum amount of daylight using this function. The maximum amount of daylight for Windsor, Ontario in 2007 is 12.18 + 3.1 sin(0.017(173.341) – 1.376) = 15.28 hours.
  • #1
livestrong136
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Help Derivatives ASAP -- maximizing sunlight through a window

3. The amount of daylight a particular location on Earth receives on a given day of the year can be modeled by a sinusoidal function. The amount of daylight that Windsor, Ontario will experience in 2007 can be modeled by the function D(t) = 12.18 + 3.1 sin(0.017t – 1.376), where t is the number of days since the start of the year.


c. The summer solstice is the day on which the maximum amount of daylight will occur. On what day of the year would this occur?

It happens when sin(0.017t – 1.376) = 1 since other constants won't change with change in t, only this sin function will change with change in t and the maximum value t hat a sine function can get is at pi/2 ie sin pi/2 = 1
i.e 0.017t-1.376 = pi/2

So how should I calculate the t from this equation

d. Verify this fact using the derivative.
Verify it by taking derivative,
d[D(t)]/dt = 3.1*0.01(cos(0.01t-1.376) = 0
Which implies that cos(0.01t-1.376) = 0
ie 0.017t-1.376 = pi/2 ..Same condition as we got in part c

My reasoning that d'(t) = 0 is correct, but rest looks wrong.

e. What is the maximum amount of daylight Windsor receives?
Maximum amount of daylight can be found out by putting t obtained from parts (c) or (d) in the equation. How would we do that.
 
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  • #2


how do you mean it looks wrong? and you said yourself sin is max at 0.017t-1.376 = pi/2, where sin is 1. so for an equation that looks like A+Bsin(Ct+F), plugging in Ct+F=pi/2 =>
D(tmax)=A+B*sin(pi/2)
solving for t won't give you an integer number of days ( but the question asks what day.. so round down)
 
  • #3


livestrong136 said:
3. The amount of daylight a particular location on Earth receives on a given day of the year can be modeled by a sinusoidal function. The amount of daylight that Windsor, Ontario will experience in 2007 can be modeled by the function D(t) = 12.18 + 3.1 sin(0.017t – 1.376), where t is the number of days since the start of the year.

c. The summer solstice is the day on which the maximum amount of daylight will occur. On what day of the year would this occur?

It happens when sin(0.017t – 1.376) = 1 since other constants won't change with change in t, only this sin function will change with change in t and the maximum value t hat a sine function can get is at pi/2 ie sin pi/2 = 1
i.e 0.017t-1.376 = pi/2

So how should I calculate the t from this equation

...
Use algebra to solve for t.

The approximate answer is 173.341.

I suppose t is an integer, so round that to the nearest integer.
 
  • #4


thnx a lot SammyS, can someone help me with d and e.
 
  • #5


For (d) What you've done seems correct. The same condition verifies the fact.

For (e) Use the 't' you get from (c) in the D(t) equation to get the maximum daylight.
 

Related to Help Derivatives - maximizing sunlight through a window

1. How do help derivatives work to maximize sunlight through a window?

Help derivatives are a type of film or coating that can be applied to windows to manipulate the way light passes through them. They work by reflecting or redirecting sunlight, allowing more of it to enter a room and reducing glare and heat. This can help to maximize natural lighting and minimize the need for artificial lighting.

2. What are the benefits of using help derivatives?

Using help derivatives can have several benefits, including reducing energy costs by decreasing the need for artificial lighting and heating or cooling systems. They can also improve the comfort and productivity of occupants by managing glare and heat, and can even help to protect furniture and other objects from fading due to excessive sunlight exposure.

3. How are help derivatives different from traditional window treatments?

Unlike traditional window treatments like curtains or blinds, help derivatives do not block out sunlight completely. Instead, they work to manipulate the way light passes through the window, allowing more natural light to enter a room. They also do not obstruct views from the window, which can be a benefit for buildings with scenic views.

4. Can help derivatives be used on all types of windows?

Yes, help derivatives can be used on most types of windows, including single and double-pane windows, as well as windows with low-E coatings. They can also be applied to both residential and commercial buildings. However, it is important to consult with a professional to ensure that the specific type of help derivative being used is suitable for the particular type of window.

5. Are there any maintenance or care requirements for help derivatives?

Help derivatives are typically low maintenance and do not require any special care. They can be cleaned with mild soap and water, and should be inspected periodically for any damage or wear. Depending on the type of help derivative used, it may need to be reapplied or replaced after a certain amount of time to maintain its effectiveness.

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