please explain answer to parts b, c: You are walking along a sidewalk toward a 40 foo

[hndalama]

You are walking along a sidewalk toward a 40 foot wide sign which is adjacent to the sidewalk and
perpendicular to it (Fig. 22).
(a) If your viewing angle  is 10 degrees, then how far are you from the
nearest corner of the sign?
(b) If your viewing angle is 10degrees and you are walking at 25 feet
per minute, then how fast is your viewing angle changing?
(c) If your viewing angle is 10 degrees and is increasing at 2degrees
per minute, then how fast are you walking?

 

[stapel]

You are walking along a sidewalk toward a 40 foot wide sign which is adjacent to the sidewalk and perpendicular to it (Fig. 22).
(a) If your viewing angle  is 10 degrees, then how far are you from the nearest corner of the sign?
(b) If your viewing angle is 10degrees and you are walking at 25 feet per minute, then how fast is your viewing angle changing?
(c) If your viewing angle is 10 degrees and is increasing at 2degrees per minute, then how fast are you walking?

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

 

[hndalama]

well for question b, I can derive the equation x=40/tany (where x= distance from sign and y =viewing angle) and from the information I know dx/dt = -25ft/min. I know the question is asking me to find dy/dt but I can't see how to derive an equation for dy/dt from the information given. when I try to isolate the angle y, I get arctan(40/x), my plan was to isolate y and then differentiate the equation but when I isolate y I get an equation with arctan in it. I don't know how to differentiate arctan because we haven't learnt that which is where I am stuck. how can I derive dy/dt from the information given?

 

[stapel]

well for question b, I can derive the equation x=40/tany (where x= distance from sign and y =viewing angle) and from the information I know dx/dt = -25ft/min.

I'm not sure why you're using this form of the relation...? Try using the usual form:

. . . . .\(\displaystyle \tan(y)\, =\, \dfrac{40}{x}\, =\, 40x^{-1}\)

I know the question is asking me to find dy/dt but I can't see how to derive an equation for dy/dt from the information given. when I try to isolate the angle y, I get arctan(40/x), my plan was to isolate y and then differentiate the equation but when I isolate y I get an equation with arctan in it. I don't know how to differentiate arctan because we haven't learnt that which is where I am stuck. how can I derive dy/dt from the information given?

If you can't use the derivative of the arc-function (because your class skipped those or whatever), try using the derivatives that your class has covered (here) and implicit differentiation with the Chain Rule.

. . . . .\(\displaystyle \sec^2(y)\, \dfrac{dy}{dt}\, =\, -40x^{-2}\, \dfrac{dx}{dt}\, =\, -\dfrac{40}{x^2}\, \dfrac{dx}{dt}\)

Use what you know, and see where it takes you!

 

[hndalama]

If you can't use the derivative of the arc-function (because your class skipped those or whatever), try using the derivatives that your class has covered (here) and implicit differentiation with the Chain Rule.

. . . . .\(\displaystyle \sec^2(y)\, \dfrac{dy}{dt}\, =\, -40x^{-2}\, \dfrac{dx}{dt}\, =\, -\dfrac{40}{x^2}\, \dfrac{dx}{dt}\)

Use what you know, and see where it takes you!

Thank you