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Problem 72 Hard Difficulty

Two vertical poles $ PQ $ and $ ST $ are secured by a rope $ PRS $ going from the top of the first pole to a point $ R $ on the ground between the poles and then to the top of the second pole as in the figure. Show that the shortest length of such a rope occurs when $ \theta_1 = \theta_2 $.

Name: two vertical poles pq and st are secured by a rope prs going from the top of the first pole to a poi
Uploaded: 2021-03-15
Duration: 3 min 56 s
Channel: Numerade
Description: Two vertical poles $ PQ $ and $ ST $ are secured by a rope $ PRS $ going from the top of the first pole to a point $ R $ on the ground between the poles and then to the top of the second pole as in the figure. Show that the shortest length of such a rope occurs when $ \theta_1 = \theta_2 $.

Answer

since $\theta,$ and $\theta_{2}$ are both acute angles, $\cos \theta_{,}=\cos \theta_{2}$ only if $\theta_{1}=\theta_{2}$

Grace H.

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

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Video Transcript

So if this problem d is going to represent the total length of the rope, so that means it's going to equal PR, which is part of the rope and R s, which is the other part of the rope. And we need to write this in terms of the angles and the heights of the polls. So what we end up getting is CO c can't of photo one is going to equal p r over our Q Coast, seeking a fatal to is going to equal r s over S T P. Q. Times Coast seeking theta one is going to equal PR so we just multiply this number of saints. That's just peek here. Here you multiply this on both sides and then similarly, we'll get the Ste. Coast SIGINT data to equals or s. So now we have that d is equal to this right here. Plus this right here. So when we differentiate that we get the the big D over D theta one is going to equal negative PQ co sick and potato one co tangent potato one minus S. T. Cosi conceded to who Tangent data to defeated two over D 3 to 1. So then we continue and we need to find an expression for this right here. We don't want to have those two differentials. So the way that we do that, as we find, uh, do you think it too? By noting that the co tangent of potato one PQ times co tangent of data one equals cure and S t co tangent faded too equals R t. So then that helps us because now we have the cutie equals P Q co tangent Say to one class S t co tangent. See that too? We differentiate this and we are able to get now that defeated two D theater one is equal to P. Q Acoustic and squared. See the one over Esti Costa Rican squared feet too. So now we can simplify our original derivative and plug this guy and right here where we were trying to replace it before. So we do that and we can simplify and we want to set this derivative equal to zero. When we simplify, we end up getting zero equals a negative co tangent, say to one coast Sea can't theater too plus co tangent stated to co Sequent say to one, then, um, we can simplify these, and we can bring this over here. So what we end up getting is a co sign of data. One equals co sign Oh, 3 to 2. Since they are both acute angles, then this only means that those are the 3 to 1 is equal to faded too, since they're both acute. And therefore the shortest length of the rope occurs when photo one equal status to

Carson M.

California Baptist University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems