Ladder rate-of-change problem

[ryankhan11]

why not do dy/dx multiplied by dx/dt ? im asking not saying thats the way to do it

[Iced0826]

I agree. It really is a classic Rate of Change problem considering I just had a very similar problem during Math class today... :\

[sam11182]

Anyone else see the beat wearing an elf hat @4:08?

(just drawn)

[alicegreenfield11]

<3

[Askate]

i love you sal!

[juju775]

THANK YOU FOR HELPING ME PASS MY CALCULUS EXAM! =D

[nike60oncore]

at: 06:15 why do you have to multiply the derivative of x^2 which is 2x by dx/dy i thought it would be just 2x + 2(dy/dx) please help!!!

[cybhunter007]

@nike60oncore

it's the chain rule: since your differentiating with respect to a different variable, you find the derivative of the x^2 (2x), followed by the prime of x with respect to the other the variable (dx/dt). If confused, try taking the derivative of x^2 with respect to x (d/dx {x^2}), what you get is 2x*dx/dx (dx/dx is x prime) and the dx/dx cancel out to become 1. the value then become equal to 2x. Hope it helps

[i8tweak]

Sal.. I hope to always have your videos available. I want to show my grandchildren your teachings! ahah Thank you.

[leaoaltamirano]

THANK YOU. YOU SAVED MY GRADE.

[atomikrobot300099]

You are a saint. You just filled in "the" missing piece of intuition for me! Thank you so much!

[93msimpson]

Me & Khan = study buddies for life

[popgirlz312]

asdfghjkl;

Sal, You're fricken amazing!

[WarMachine425]

ur either really good with writing with a mouse, or ur using a pen

[BarbierNicholas]

This is a stupid question:

but, after .5 seconds wouldn't the bottom point have moved over .5*4=2m; 2+8=10m so the ladder would be laying on the ground meaning that the y value had moved 6m in .5 seconds? that would imply 12 m/s (technically or -12, whatever since it's velocity) yet this video tells me 16/3 or -16/3 m/s. Where does my intuition go wrong?

[yousmath]

@BarbierNicholas

it confused me to

[Jtwaz1]

@BarbierNicholas

That's the average speed it's falling from the interval x=[8-10], not the speed it's falling at a given point.

Think about it like this, you go out driving, starting at 0 mph, and end up going 60 miles in one hour, giving you an average speed of 60 mph's, but you didn't start at 60mph, you started at 0 mph, so the average speed isn't necessarily the speed at any given point.

and that's where Calculus starts to diverges from Algebra

[sweentnhot0104]

@BarbierNicholas This might be a distant analogy, but think about it this way. The car is moving in a constant velocity, meaning it IS moving. However, the acceleration would be a zero since there is no change in the velocity. However, similar to this problem, this is not always the case, because normally the acceleration changes (the car does not move in a constant speed). That's when we need calculus to figure out the rate of change

[abhi86255]

thank u...........

[Flows234]

I was working on this for days! and it just clicks after watching your vids! Thanks Salman

[SpiderPig2494]

Y U NO BE MY TEACHER ?!?

[demonangels67]

really clear

[s200960170]

You make life so easy sal.. :) thank you sooo much..

[thieflord08]

lol after i finished this i just cussed my self out for being so stupid and not getting this in class.. thanks fly man!

[coffeecuppepsi]

Thank you Sal

[elquijote85]

you sir are allowed to teach. cant say the same for my teacher.

[1001014]

I have this question sal: I think you only found the rate of change at that instant and not a general expression. Sorry if I'm wrong, but I found: x=8+4t so y=sqrt(10^2-x^2) and y'=[8(t-2)/sqrt(4t^2-16t+9)] and when t=0, y'=-5.33, the same result as you. But erm as plainly obvious y' cannot be a constant. I think by substituting 6 and 8 into your implicitly done y' you chose to find the rate of change of ladder vertically at exactly t=0 moment. Love you Sal btw you're a great teacher!

[Ghiltzisme]

You are my Math Hero!

[rayhanna90]

great job thanks.......you awsome...but i have a suggestion and that is if you could number you videos or make one topic connect to each other...ta:)

[sweetpnayxangel]

thank you so much!! this really helped

[strat0caster266]

Sal, you are my lord and savior. Thanks to you I have successfully tacked my first related rates problem. I now feel confident heading into my test tomorrow. Keep up the great work, and know that we all appreciate it.

[Boofyable]

KHAN!!!!!!!!!!!!!!!!!! you are the greatest!

[classifired]

amazing,,thanks! :D

[justapurpose2]

No....no....no..... the classic rate of change problem is more like the one dealing with shadows....

[yas72090]

greased up haha

[alliem3131]

solid sal you got me through physics and now calc

[jomorrissey07]

that kicked ass, thank you

[roger9kuo]

thanks. i had a similar problem just that i was solving for dx/dt instead of dy/dt

[tazdevil112]

LOL

[roger9kuo]

hows that funny?

[cassi1126]

Thanks, this was a great help~

[yoliz]

thanks, this helped me a lot.

[USMChiLD]

thanks again!

[mattkelll]

Hi...thanks for all the great tutorials. Could you show an inscribed triangle rate of change problem? Thanks.

[HorsemenofFury]

i bet your ladder is 10 meters long.

[pacmanty]

thanks =D

[masood365]

hi i love watching ur tutorials. I have the same question but little different if you could help me. ( an extension lader with the top resting on a vertical wall is being extended at a rate of 10 m/min. the base is 5 m from the wall. find the rate of sliding up of the top of the lader when the lader is 13 m long? thanks

[Jechux1106]

thnx a lot its really clear now.

[elfangor3]

Thank you for doing this and explaining it so well.

[dottorecannelle]

bad ass. real cool.

[GoldAK47]

great vid...