Comparing Daylight Increase in Philadelphia Using a Model

[lexismone]

1. Homework Statement

A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is given.
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on June 5 and June 15. (Round your answers to four decimal places).

2. Homework Equations

L(t)=12+2.8sin[(2pi/365)(t-80)]

3. The Attempt at a Solution

i have no clue were to start. or what to do, please help.

i tried
L(t)=12+2.8sin[(2pi/365)(5-80)]=9.3088
L(t)=12+2.8sin[(2pi/365)(15-80)]= 9.481

it seemed to easy to be right, and of course it was. please help?

 

[Mark44]

1. Homework Statement

A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is given.
Use this model to compare how the number of hours of daylight is increasing in Philadelphia on June 5 and June 15. (Round your answers to four decimal places).

2. Homework Equations

L(t)=12+2.8sin[(2pi/365)(t-80)]

3. The Attempt at a Solution

i have no clue were to start. or what to do, please help.

i tried
L(t)=12+2.8sin[(2pi/365)(5-80)]=9.3088

This is L(5). L(t) is what you have in Relevant equations.

L(t)=12+2.8sin[(2pi/365)(15-80)]= 9.481

This is L(15).

it seemed to easy to be right, and of course it was. please help?

What they're asking about is the slope of the tangent line at a) (5, L(5)) and b) (15, L(15)).

For the first one, calculate the slope of a secant line for two points on the graph of your function. The two points should straddle (5, L(5)). The closer together the two points are, the better your approximation will be.

Do the same for the other point at (15, L(15)).

 

[lexismone]

okay, so
a) (5, 9.3088)
for the slope, should i use a point to the left and right of 5, such as 4.99 and 5.01?

b) (15, 9.481)
and the same for this part, 14.99 and 15.01?

 

[Mark44]

Right, those would be good choices.

 

[lexismone]

a) (4.99,9.3087) (5.01,9.309)=(9.309-9.3087)/(5.01-4.99)= .015

b) (14.99,9.4808)(15.01,9.4812)= (9.4812-9.4808)/(15.01-14.99)= .02

so that's my answer?

 

[Mark44]

a) (4.99,9.3087) (5.01,9.309)=(9.309-9.3087)/(5.01-4.99)= .015

Let's clean this up.
The slope of the secant line between (4.99,9.3087) and (5.01,9.309) is
(9.309-9.3087)/(5.01-4.99)= .015

Unfortunately, you aren't evaluating your function correctly, something I didn't notice before.

A model for the length of daylight (in hours) in Philadelphia on the tth day of the year is given.

L(5) represents the number of hours of daylight on Jan 5. You're going to have to count days to see which days of the year June 5 and June 15 are.

b) (14.99,9.4808)(15.01,9.4812)= (9.4812-9.4808)/(15.01-14.99)= .02

so that's my answer?

 

[Ray Vickson]

okay, so
a) (5, 9.3088)
for the slope, should i use a point to the left and right of 5, such as 4.99 and 5.01?

b) (15, 9.481)
and the same for this part, 14.99 and 15.01?

Since this is the "Calculus and Beyond" Forum, my first suggestion would be to use exact derivatives. If you don't know what derivatives are or how to compute them, then we have a different discussion!

RGV

 

[Mark44]

Since this is the "Calculus and Beyond" Forum, my first suggestion would be to use exact derivatives. If you don't know what derivatives are or how to compute them, then we have a different discussion!

That's probably good advice. I was operating under the assumption (probably unwarranted) that the OP had not yet been exposed to differentiation.