0:00 A kite 100 ft. above the ground moves horizontally at a speed of 8 ft/s 0:57 Draw a sketch 4:17 Come up with your equation 7:54 Implicit Differentiation 10:04 Solve For the Desired Rate of Change
Very thorough (and absolutely correct) but you didn't need to slog through any of it. dθ/dt = dθ/dx · dx/dt. In this case, θ = arccot(x/100). d[cot(θ)]/dθ = - csc^2(θ) = - (200/100)^2 = - 4. By the inverse function theorem then dθ/dx = 1/100 × 1/( - csc^2(θ)) = 1/100 × ( - 1/4) = - 1/400. Therefore, dθ/dt = - 1/400 × 8/1 = - 1/50 radians per second. ◼
0:00 A kite 100 ft. above the ground moves horizontally at a speed of 8 ft/s
0:57 Draw a sketch
4:17 Come up with your equation
7:54 Implicit Differentiation
10:04 Solve For the Desired Rate of Change
my friend you just got another subscriber ... searched and searched and you broke it down . major thanks
Glad I could help!
Very thorough (and absolutely correct) but you didn't need to slog through any of it.
dθ/dt = dθ/dx · dx/dt.
In this case, θ = arccot(x/100).
d[cot(θ)]/dθ = - csc^2(θ) = - (200/100)^2 = - 4.
By the inverse function theorem then dθ/dx = 1/100 × 1/( - csc^2(θ)) = 1/100 × ( - 1/4) = - 1/400.
Therefore, dθ/dt = - 1/400 × 8/1 = - 1/50 radians per second. ◼
Thanks for your help!
Of course!
helping me learn for my exam next week, thanks!
You’re welcome, I’m glad it helped!
very helpful! thank you so much
You’re welcome, I’m so glad it helped!
So helpful thank you!!
You’re welcome, happy to help!
Nice work.
Thank you!
good vid!
Great
Thank you!
Please can you show how you got -1/50 I got 2
maybe use the calculator correctly..
You the man!
Thank you!