Related Rates - Angle of Elevation Problem

[@TheOrganicChemistryTutor]

Next Video: kzbin.info/www/bejne/iamVnYOQodOKnNU Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections

[@claricebryant7798]

I was going crazy with a problem like this. Thank you so much for making it understandable!

[@georgesadler7830]

MR. Organic Chemistry Tutor, thank you for an incredible video/lecture of The Angle of Elevation in the Related Rates section of Calculus One. A solid background in Trigonometry really helps with understanding Related Rates and also drawing a picture of the problem. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

[@BillyBob-tr5cv]

This is helpful and all...but I seem to not be able to find any videos over finding the actual angels of elevation...all I wanted to do was to make sure I was doing it right 😭

[@TheFartastic1]

same

[@rachelhiltystudenthbhs2575]

the hero we all needed

[@IamKudos]

isnt the final answer 125 degrees since you started with 60 degrees?

[@knhsrvknhsrv5801]

Peace,No its rads because 60 is the same as pi over three and that would not intervene in the solution.

[@orhionsefeighile620]

That's was a great 👍 video I ever had. Thanks for sharing

[@davidstevensom4330]

I would really like to know why the answer is in radians and not degrees. Please help me out. Thanks. David.

[@chrisp2639]

Because theta is measured in radians

[@Kriffle]

@@chrisp2639 but theta is given to us in degrees, so wouldn't it be in degrees?

[@chrisp2639]

@@Kriffle My previous response was not very clear at all, apologies. Rate of change problems almost always involve taking the derivative of θ (theta) with respect to time (t). The derivatives of the trigonometric functions are derived using their definitions in terms of radians. A radian is defined in terms of arc length. Specifically, an angle of 1 radian subtends an arc equal in length to the radius of the circle. This connection between angles and arc lengths makes calculations involving rates of change more natural in radians. Lastly, for simplicities sake, working in terms of radians will remove the additional π/180 factor (or its reciprocal). If the problem requires it (which I have never seen myself), you can always reintroduce the π/180 factor (or its reciprocal) to convert back but this would not be practical as it would just increase complexity. I hope this is helpful. Related rates were one of the two most challenging topics for me in Calculus 1 and the only way I got through it is by watching a BUNCH of examples and essentially memorizing them. Good luck!

[@boredash4020]

@@chrisp2639 what i don't understand is that all throughout our working we used theta as 60 degrees,so how could the answer be in radians? or did we assume that the question was supposed to be given in radians?

[@parvatin1759]

​@@boredash4020 this is probably too late, but i hope it helps anyone else reading this. yes, theta was given to us in degrees, and you could go about solving the problem in degrees and get 125 as the final answer (disregarding the units of the answer). but even if you convert 60 degrees (the given value of theta in this case) into radians and solve the problem in radians, you will still end up with 125. going step by step, at 3:36 we have tan(60 degrees), which is tan(pi/3) when converted to radians. either one will result in sqrt(3). therefore, it does not matter whether the plane's angle of elevation is given to us in degrees or radians. like ​ @chrisp2639 said, the result will involve radians because the derivative of trig functions as we know them (for example, in this case it is d/dx(tanx)=sec^2x) are calculated in terms of radians. when we say that the derivative of tan(x) is sec^2(x), you are assuming that x is in terms of radians because there is no unit listed. it says tan(x), not tan(x°). some degree-based trig derivatives, for example, are d/dx(sin(x°))=(π/180)cos(x°) and d/dx(cos(x°))=-(π/180)sin(x°). knowing the standard trig derivatives in terms of radians, hopefully you can see some kind of pattern happening there. if you'd like to look further into this subject, try looking up something along the lines of "trig derivatives in terms of degrees". hope that helped :)

[@SoupTurtle16]

I have a question when u did d/dt of the whole equation why had y stayed the same shoould it been chnaged to dy/dt or does it just remains y

[@habtamu77]

You have given y=3 so it is not changing then you can write 3/x as 3 (1/x) =3 d/dx (1/x) take the constant in the front and differentiate what is changing

[@alexfish7792]

Did you skip the product rule? Since both y and x are functions of t, isn't it really saying f(t)*g(t)? Did you skip it because you knew dy/dt was 0 already or instead were you saying y(x^-1) really means 3(x^-1)? (meaning no product rule since is 3 is a constant)? Please let me know if I'm missing something. You're an amazing teacher by the way.

[@VeryBigDave]

I'm not sure if you figured it out or if it still matters to you but since y was a constant, you don't have to worry about the product rule. You can simply take it out before deriving just like you would with a number. Since y = 3, y acts as 3 in that part of the equation.

[@alexfish7792]

@@VeryBigDave Yeah that's exactly what I was thinking I just didn't notice it since he kept it as "y" instead of explicitly writing 3. Thanks for clarifying! :)

[@Badsniperarmy]

good stuff man thank you

[@jaehoahn4810]

when you find derivative of your equation you did not add x^-1(dy/dt). Why?

[@alexfish7792]

I think because he already knew dy/dt = 0. I also was wondering about this. Or maybe there is no "y as a function of t" since we already know it as 3? Maybe y(x^-1) really means 3(x^-1)?

[@DaiMoscv]

Because y is constant 3 and when you differentiate a constant it's always 0

[@alyssamelton4058]

Thank you!

[@johnnolen8338]

One might guess that radians per hour are the correct units to express dθ/dt, but 125 radians per hour is such a large value that it is essentially meaningless; esp. considering that the airplane closes the distance to the observer in less than 12.5 seconds.

[@danielaramos5880]

I might be wrong but shouldn't the speed be positive no matter where the plane is going? Speed is a scalar quantity, it doesn't have direction; it would only make sense to apply direction if it was a vector, like velocity. Just a thought :)

[@alexfish7792]

An observer sees the plane flying "toward him". This gives the plane's speed a direction making it a vector. :)

[@UnlikelyToRemember]

you can choose any convenient coordinate system, including one where the dx/dt is positive.

[@zubair1424]

Why is Y not zeroed out since it is a constant?

[@knhsrvknhsrv5801]

Peace,If you want to do that you would have to do the product rule try it and you'll know why.

[@1polyron1]

absolute unit

[@albertelectron3961]

Thanks 🙏

[@davidstevensom4330]

You messed up the final answer. It's 125 degrees per hour. Since you took the cosine of 60 degrees squared, you were working with degrees. Not radians.

[@knhsrvknhsrv5801]

Peace , its in rads actually because 60 is the same as pi over three so that wouldn't change the solution same answer

[@davidlu5904]

Couldn't you just have done: d/dt(tan θ = 3x^-1) sec^2(θ) * dθ/dt = -3x^2 * dx/dt x=sqrt(3) and dx/dt = -500, so sec^2(θ) * dθ/dt = 500. θ = 60 degrees, so sec^2(60) = 1/cos^2(60) (into calculator = 4) so 4 * dθ/dt = 500, dθ/dt = 500/4 = 125 rad/hour Or did I do something wrong?

[@SheLxvDraco]

I'm so confused with this stuff

[@joshuamishaelmacapobre6396]

hahah dude I know this is late but you also helped me with this prob

[@leonabyunparkii9225]

I just wanna ask if you can do it via the sin instead of the tan?

[@endvine9951]

can you?

[@smozaf]

can anyone explain to me how the unit rad appeared in the final answer? I get that it's the angle's unit but I'm just trying to understand when and where rad appears in the calculations and not just as the final answer. TIA

[@beanboy1821]

i believe it's just a simple conversion formula: (x° × π/180)

[@fcantil]

@@beanboy1821 you say that but there's no conversion from degrees to radians happening anywhere in the video at all. i'm so confused. there's a couple of other comments also pointing this out, but there are either no replies or just ones that don't make sense.

[@shalekthewise]

@@fcantil I’m not exactly sure why but I’m taking calculus rn and radians is always the unit of measure in these sorts of problems

[@fcantil]

@@shalekthewise I figured it out and I forgot to edit my comment. When solving for the tangent of an angle in degrees, it is first converted to radians, since radians are unitless. He basically took a shortcut and didn't explain that step. That's why it ends up becoming radians in the end.

[@golddddus]

Wrong. 11:35 sec(theta)^2 (d(theta)/dt)=-Y*X^(-2)*d(theta)/dt is OK but then Theta is not constant.( sec(theta))^2 =1 + tan (theta)^2= 1+Y^2/X^2 So d(theta)/dt= -(1+Y^2/X^2)^(-1) *Y/X^2*(dx/dt)=-Y/(X^2+Y^2)*(dx/dt)= 3/(x^2+ 3^2)*(-500)=1500/(x^2+9) rad/h Your answer 125 rad/h is meaningless. 125 rad is about 120/6= 20* 360 degree. From your picture is clear that in infinity d(theta)/dt = 0. And maximum theta is Pi rad.

[@alexthegoatyt5207]

This was your first miss tbh

[@sylshooten410]

now do a rocket