A STREET LIGHT IS MOUNTED AT THE TOP OF A 15-FT-TALL POLE. A MAN 6 FT TALL WALKS AWAY FROM THE POLE
Vložit
- čas přidán 17. 07. 2024
- A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?
This related rates triangle problem is a good example of a related rates calculus word problem. Even if you don't have this specific problem required for homework, this related rates triangle problem is a good demonstration of techniques that can be used in other related rates problems dealing with related rates and other geometric shapes. That's because it can be solved using the same 4 step process as all other related rates calculus word problems:
0:00 A street light is mounted at the top of a 15-ft-tall pole
0:32 Draw a sketch
2:40 Come up with your equation
9:58 Implicit Differentiation
10:45 Solve For the Desired Rate of Change
WATCH NEXT
+ More Related Rates Examples - • Related Rates
+ Implicit Differentiation - • Implicit Differentiati...
READ MORE
+ Chain Rule - jakesmathlessons.com/derivati...
+ Related Rates - jakesmathlessons.com/derivati...
+ Implicit Differentiation - jakesmathlessons.com/derivati...
YOU MIGHT ALSO BE INTERESTED IN...
+ Download my FREE calc 1 study guide - jakesmathlessons.com/Calculus...
+ My Complete Calculus 1 Package - jakesmathlessons.com/complete...
+ Work with me! - Come to Wyzant for some 1 on 1 online tutoring with me and get a $40 tutoring credit for FREE - is.gd/cqeuOv
+ Problem credit: Chapter 4.1 #14 in Single Variable Calculus: Concepts and Contexts by James Stewart - amzn.to/2ChukmL
Some links in this video description may be affiliate links meaning I would get a commission for your purchase at no additional cost to you.
0:00 A street light is mounted at the top of a 15-ft-tall pole
0:32 Draw a sketch
1:37 Come up with your equation
9:58 Implicit Differentiation
10:45 Solve For the Desired Rate of Change
Taking differential calculus is hard working full time and not having lectures. You have helped me get a better understanding faster. Thank you.
That’s awesome, I’m happy to hear my videos helped. That is a tough thing to do. Keep up the good work!
11 min long video = 1 hr long math lecture
Thank you, hope it helped!
AM LITERALLY CRYING RIGHT NOW. you made it look sooo simple. THANK YOU SOOO MUCH. i just subbed
Awesome! I’m glad it helped so much!
dude thank you so much! your videos are really helping me understand the concepts that my book is teaching. Way better than my professor lol
Awesome! I’m glad you’re finding them so helpful!
That was so easy to follow ❤ thanks!
Very helpful thank you!
hey man, we did a pop quiz in the middle of our class and you just saved my grade, have a good one
Awesome! I’m glad it helped!
Awesome video! Thanks for the help!
Nick Weber you’re welcome!
Really well explained man, keep it up!!
I'm glad it helped!
Thank you for this! The music was a nice touch.
Thank you! I’m glad you like it. I really liked the addition of some background music too!
amazing explanation
Thank you so much!
You’re welcome!
Great presentation. Bravo keep up the great work!
Thank you, I will!
thank you so much man
You’re welcome!
thanks bro you helped a lot
You’re welcome, happy to hear that!
why do we take it from all the part of the triangle like the x and y and not just the y part
?
thank you
You’re welcome!
You solve my calc hw, tysm
You’re welcome!
Thank you very much for this video. It explained clear. But I have one question: The speed of shadow (dZ/dt) = 25/3 (ft/sec) is a constant. Shouldn't the speed of the tip of the shadow moving faster and faster as the man is moving further away from the pole, because the incline angle from the tip of shadow to the light gets smaller and smaller? So the speed of the shadow dX/dt should not be a constant. Right now, X = Z-Y. dX/dt = dZ/dt - dY/dt = 25/3 - 5 = 10/3 (ft/sec), which is a constant. Please clarify for me. Thank you very much.
It does seem like that might be the case, but the math confirms it is actually a constant speed. You could test this too by taking a few points in time and drawing what those triangles would look like. You would see that the tip of the shadow moves the same distance between t=1 and t=2 as it does between t=10 and t=11, or any other two time values that are the same distance in time from each other. Therefore its speed is constant.
Great video!!, Can I have the music name around 6:00 ? I really like the music
Honestly, I can't remember which song it is unfortunately. I got it on epidemicsound.com though.
why is dy/dt 5? is that is x
So, no use of y=40 then?
why couldnt we just replace the y variable with 40?
You can’t plug in 40 for y before the implicit differentiation step because it’s changing over time. Any variables that are changing over time can’t be plugged in until after the implicit differentiation step. You can plug in constants, but y changes as the person walks, so it’s not a constant.
should remove the back ground music next time
Thank you for the feedback, I appreciate that. Did you find the music distracting? Was it just too loud?
Jake's Math Lessons I found it a little bit distracting
Thanks for the heads up!
@@JakesMathLessons you did a great job explaining though!!
@@saeroyipark6047 thank you!