Related Rates With Right Angle Trigonometry (Kite Example)

Related Rates With Right Angle Trigonometry (Kite Example)

[@dreiiii_0182]

I can't thank you enough Mr. Barnes! your method is so much easier than the one I learned. Thank you always for helping

[@mathwithmrbarnes]

Glad I could help!

[@johnnolen8338]

You solved the problem correctly, well done. However, you did it the hard way in my opinion. It is worthy of note that cot(θ) = x/40. Differentiating both sides of this equation with respect to θ gives - csc^2(θ) = 1/40·dx/dθ --> - (80/40)^2 = 1/40·dx/dθ.
Taking the reciprocal of this expression gives: - 1/4 = 40·dθ/dx. Therefore, dθ/dx = - 1/160.
By the chain rule dθ/dt = dθ/dx·dx/dt, it follows that dθ/dt = (- 1/160)(3/1) = - 3/160 rad/s ◼

[@derickflorian4724]

Thank you Mr. Barnes! I had just one question. Doesn't the x^2 from the denominator and the one from the numerator cancel out? In that case you could skip computing the value of x, right? Thanks, for related rates I would be lost without youtube :).

[@adhdHD09]

THANK YOU SO MUCH. I'VE GOT THE SAME EXACT QUESTION, EXCEPT FOR THE GIVEN VALUES OF COURSE

[@johnsturm4407]

At 1:44 shouldn’t that be dx/dt?

[@clariguycameron]

Yes

[@jeffersonlagua]

kite, at a height of 42 ft is moving horizontally at a rate of 5 ft/s away from the boy whoflies it. How fast is the cord being released when there are 63.7809 ft out?

[@jean-philippelopezletz9948]

Good video but confused the shit out of me when you put dx/dy=3 m/sec instead it's dx/dt

[@Aviator.86]

Thank you for this video! It helped me a lot.

[@surendrakverma555]

Very good 🙏🙏🙏🙏🙏

[@iishadowii7477]

Thank You!