Related Rates - Angle of Elevation Problem

This calculus video tutorial on application of derivatives explains how to solve the angle of elevation problem in related rates.

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Introduction to Limits:

Derivatives - Fast Review:

Introduction to Related Rates:

Derivative Notations:

Related Rates - The Cube:

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Inflated Balloon & Melting Snowball:

Gravel Dumped Into Conical Tank:

Related Rates - Area of a Triangle:

Related Rates - The Ladder Problem:

Related Rates - The Distance Problem:

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Related Rates - Airplane Problems:

Related Rates - The Shadow Problem:

Related Rates - The Baseball Diamond Problem:

Related Rates - The Angle of Elevation Problem:

Related Rates - More Practice Problems:

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Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

[claricebryant]

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

[georgesadler]

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 YouTube TV with the Organic Chemistry Tutor.

[BillyBob-trcv]

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 😭

[IamKudos]

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

[davidstevensom]

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

[orhionsefeighile]

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

[johnnolen]

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.

[jaehoahn]

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

[SoupTurtle]

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

[alexfish]

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.

[zubair]

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

[davidstevensom]

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.

[danielaramos]

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 :)

[davidlu]

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?

[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

[leonabyunparkii]

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

[golddddus]

Wrong. 11:35 sec(theta)^2 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) 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.