Related Rates - Angle of Elevation Problem
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- This calculus video tutorial on application of derivatives explains how to solve the angle of elevation problem in related rates.
Introduction to Limits: • Calculus 1 - Introduct...
Derivatives - Fast Review:
• Calculus 1 - Derivatives
Introduction to Related Rates:
• Introduction to Relate...
Derivative Notations:
• dy/dx, d/dx, and dy/dt...
Related Rates - The Cube:
• Related Rate Problems ...
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Inflated Balloon & Melting Snowball:
• Related Rates - Inflat...
Gravel Dumped Into Conical Tank:
• Related Rates - Gravel...
Related Rates - Area of a Triangle:
• Related Rates - Area o...
Related Rates - The Ladder Problem:
• Related Rates - The La...
Related Rates - The Distance Problem:
• Related Rates - Distan...
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Related Rates - Airplane Problems:
• Related Rates - Airpla...
Related Rates - The Shadow Problem:
• Related Rates - The Sh...
Related Rates - The Baseball Diamond Problem:
• Related Rates - The Ba...
Related Rates - The Angle of Elevation Problem:
• Related Rates - Angle ...
Related Rates - More Practice Problems:
• Related Rates - Conica...
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I was going crazy with a problem like this. Thank you so much for making it understandable!
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 😭
same
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 TH-cam TV with the Organic Chemistry Tutor.
the hero we all needed
isnt the final answer 125 degrees since you started with 60 degrees?
Peace,No its rads because 60 is the same as pi over three and that would not intervene in the solution.
That's was a great 👍 video I ever had. Thanks for sharing
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
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
I would really like to know why the answer is in radians and not degrees. Please help me out. Thanks. David.
Because theta is measured in radians
@@chrisp2639 but theta is given to us in degrees, so wouldn't it be in degrees?
@@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!
@@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?
when you find derivative of your equation you did not add x^-1(dy/dt). Why?
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)?
Because y is constant 3 and when you differentiate a constant it's always 0
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.
good stuff man thank you
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.
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.
@@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! :)
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.
Peace , its in rads actually because 60 is the same as pi over three so that wouldn't change the solution same answer
Why is Y not zeroed out since it is a constant?
Peace,If you want to do that you would have to do the product rule try it and you'll know why.
Thank you!
Thanks 🙏
absolute unit
I just wanna ask if you can do it via the sin instead of the tan?
can you?
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 :)
An observer sees the plane flying "toward him". This gives the plane's speed a direction making it a vector. :)
you can choose any convenient coordinate system, including one where the dx/dt is positive.
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?
I'm so confused with this stuff
hahah dude I know this is late but you also helped me with this prob
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
i believe it's just a simple conversion formula: (x° × π/180)
@@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.
@@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
@@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.
This was your first miss tbh
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.
now do a rocket