Calculus Related Rates - The Shadow Problem

This video show how to find the rate of change of the tip of a shadow from a light post.

[@TojiZynin]

This channel is amazing thank you this is helping me study for finals

[@crowsmathclass]

Thank you. Glad it is helping.

[@luvniwa]

that into gave me whiplash but this was very helpful I think idk I haven't watched the video yet but I think it will be

[@crowsmathclass]

👍

[@hmmm1482]

I've noticed horizontal triangles and found that velocity should be (15/6)*(dx/dt), what is the mistake?

[@vymbz2653]

So is x=40 not used because the derivative is linear?

[@crowsmathclass]

It wants to know at 40

[@tishnn]

ive watched this 4 times and i cant get it. its not you, its me, I promise.

[@yaanbutdiff6693]

Thank you sir a lot sir for this video i really appreciate it ❤️❤️❤️❤️

[@crowsmathclass]

You’re welcome.

[@simplyydee]

great video

[@crowsmathclass]

Thank you.

[@ignacywozniak3373]

Love the voice

[@crowsmathclass]

Thank you.

[@reecemoulton6709]

Thank you sir!

[@crowsmathclass]

You’re welcome

[@user-zg8bi8ss5u]

Where did you get the 5/3

[@tonynguyen9755]

2/3 +1 = 5/3

[@rodolfodumaliangiii7342]

thank you!

[@crowsmathclass]

You’re welcome.

[@matteodhoe]

thank you

[@crowsmathclass]

You’re welcome.

[@user-bi9ee9xe2m]

Thanks....

[@crowsmathclass]

You’re welcome.

[@cuckoo46]

Sir, your method is correct but answer is wrong it should have been 10/3ft/sec . not 25/3ft/sec. kindly check the mistake you have made taking the derivative of (x+ 2/3x) it should read 2/3dX/dt not 5/3dx/dt

[@khiensilog]

The derivative should be (1+2/3)dx/dt, which is 5/3 dx/dt

[@uuq1193]

nah add dx/dt and dy/dt