Modeling the length of a plank as a function of an indicated angle

Holydog23

New member
Joined
Aug 16, 2018
Messages
8
D402D06C-DB9C-4E31-870A-C1D918FDC7CE.jpg
This a problem in the functions section of my calculus textbook and I was having massive trouble with setting up this function. I have tried to use sin and cosine functions to isolate the hypotenuse which is the plank in this context but I always end up with another variable besides theta because I don’t know the lengths of the triangles sides, only parts such as 3 and 4 ft. The final answer is supposed to be5CD69350-D56F-42AD-8549-6B4FA8872FBB.jpg

and I honestly can only slightly get 3csc(theta) but I’m getting another variable that I can’t rid of. I’m making up these variables because I have to subtract or add unknown lengths to get those trig ratios
 
View attachment 9961
This a problem in the functions section of my calculus textbook and I was having massive trouble with setting up this function. I have tried to use sin and cosine functions to isolate the hypotenuse which is the plank in this context but I always end up with another variable besides theta because I don’t know the lengths of the triangles sides, only parts such as 3 and 4 ft. The final answer is supposed to beView attachment 9962

and I honestly can only slightly get 3csc(theta) but I’m getting another variable that I can’t rid of. I’m making up these variables because I have to subtract or add unknown lengths to get those trig ratios
Please share your work with us showing - what you get and how do you get that.

By the way, I am getting the posted answer.
 
F7E45B82-1A52-4D07-BE73-417B3F685C10.jpg
Here’s all I have so far. Those are the insights I’ve collected off the diagram. I do not know anything else beyond this. I’ve exhausted a great deal of effort trying to somehow get the final answer. I’m not even sure if introducing those variables was necessary or not. It’s just something I learned to do as a problem solver
 

Attachments

  • 9592B8AB-36DF-409B-B55F-AFA03998D65E.jpg
    9592B8AB-36DF-409B-B55F-AFA03998D65E.jpg
    8.5 KB · Views: 1
View attachment 9961
This a problem in the functions section of my calculus textbook and I was having massive trouble with setting up this function. I have tried to use sin and cosine functions to isolate the hypotenuse which is the plank in this context but I always end up with another variable besides theta because I don’t know the lengths of the triangles sides, only parts such as 3 and 4 ft. The final answer is supposed to beView attachment 9962

and I honestly can only slightly get 3csc(theta) but I’m getting another variable that I can’t rid of. I’m making up these variables because I have to subtract or add unknown lengths to get those trig ratios

what is the other variable you can't get rid of?
It may help to write the equation in terms of sin and cos.
i.e. L=3/sin + 4/cos
then convert it knowing 1/sin=csc, and 1/cos=sec.

the relations for sin and cos are probably easier to see this way;
sin(theta)=3/L1; cos(theta)=4/L2
at any rate you get expressions for L1 and L2 and add them.
L(theta)=3csc(theta)+4sec(theta)
 
Please share your work with us showing - what you get and how do you get that.

By the way, I am getting the posted answer.
attachment.php

Assume that
the plank touches the wall at W

the plank touches the ground at G

The length of the plank = L

cot(θ) = [Lcos(θ) - 4]/3

Simplify....

Please let us know if you do not understand any term or step.
 
Please share your work with us showing - what you get and how do you get that.

By the way, I am getting the posted answer.

attachment.php

Assume that
the plank touches the wall at W

the plank touches the ground at G

The length of the plank = L

cot(θ) = [Lcos(θ) - 4]/3

Simplify....

Please let us know if you do not understand any term or step.



Where did you derive the cotangent relationship from? I can’t see how I was supposed to figure that out from the diagram?
 
what is the other variable you can't get rid of?
It may help to write the equation in terms of sin and cos.
i.e. L=3/sin + 4/cos
then convert it knowing 1/sin=csc, and 1/cos=sec.

the relations for sin and cos are probably easier to see this way;
sin(theta)=3/L1; cos(theta)=4/L2
at any rate you get expressions for L1 and L2 and add them.
L(theta)=3csc(theta)+4sec(theta)

How did you get that cos(theta) is 4 divided by the second piece of the plank? Don’t you have to account for the initial distance between the point where and plank touches the ground and the middle of the the sawhorse? I understand that between the sawhorse and the wall there is 4ft but what about before then?
 
How did you get that cos(theta) is 4 divided by the second piece of the plank? Don’t you have to account for the initial distance between the point where and plank touches the ground and the middle of the the sawhorse? I understand that between the sawhorse and the wall there is 4ft but what about before then?

there is an angle of theta between the plank and the horizontal, regardless of where you measure it.
there is 4 ft horizontal distance between the saw horse and the wall.
therefore cos(theta) =4/L2.

do you see that the angle between the plank and the horizontal is the same at the saw horse as it is where the plank touches the ground?
it may help to; ignore what is to the left of the saw horse, and Imagine the saw horse is the ground.
then you can use the L2 part of the plank, and its horizontal component 4, to define the cos.

If you were to include the 'initial' distance (horiz. between the ground and saw horse) with the distance 4, you would then have to use the total length of the plank in defining cos (theta).
 
Last edited:
attachment.php

Assume that
the plank touches the wall at W

the plank touches the ground at G

The length of the plank = L

The wall touches the ground at W' →

GW' = L*cos(Θ)

The saw-horse touches the planck at P

Perpendicular from P to the ground is P'

then for triangle PGP'

cot(Θ) = (base)/(perpendicular) = GP'/PP' = (GW' - 4)/3

cot(θ) = [Lcos(θ) - 4]/3

Simplify....

Please let us know if you do not understand any term or step.
 
Top