@pyepyo1 This is just a rephrasing of the formula. If D=R*T and you want to solve for rate, you have to isolate R. The way you do that is to divide both sides by T, R*T/T = D/T, the T’s on the side with R divide out to be 1 which leaves you with R=D/T. If you plug in distance and time into this you can easily calculate the rate.
I freakin love you, no seriously I do. I'm taking Calc I next semester and just by watching your videos and looking over my textbook, I'm not so terrified anymore.
after i watched your videos in precalculus and calculus you helped me a lot to understand what is going on in calculus books and what my prof is saying and i got (A) IN CALCULUS and im now good in physics thank you very much thinkwell and edward burger .... [ thinkwell = :) + :) ]
@pyepyo1 Not sure what exactly you are asking here, but as long as you have distance and time, you can calculate rate from the D=R*T formula. Glad you enjoy our videos and hope we are able to help you jump ahead!
Yes, in this case the answer can be found with simple algebra. He used a very simple example. Calculus comes into play when you have a position function (for example I will use y=t^2. y being position and t being time.) Obviously, this has you changing speeds so it would be considerably harder to figure out some algebra to find his speed at say t=20 (if it is even possible at all). Calculus tells us he is going 40 miles / hour at t=20 if position is in miles. I hope you enjoy Calc it is
An instant is an infinitely small portion of a second. However small a non-zero portion you come up with, an instant is smaller. So, nothing actually happens during an instant, because anything that takes time takes more time than an instant. However, because motion is continuous, the bike is still moving during an instant, even though it can't accomplish anything. So, the bike has a speed it travels during that instant, but the instant is so short that it covers zero distance in an instant.
Hey i been tol i schmelll tooo? anyway..perhaps a Plank time measurement 10 to the minus 33rd power of magnitude? or the size of a single string as in string theorem..or most flavors of it anyway..
@ThinkwellVids so you mean, you get the rate from the question ? correct? i'm just a middle school student who want to jump a head a bit.so tyvm.. ur videos are awesome. :D
That was his average velocity. We're talking about instantaneous velocity. He could have been speeding up or slowing down during the bike ride, so the average velocity is not necessarily his instantaneous velocity at the 20 mile point.
i don't get the part where you said { D,R,T ; D=R*T ; how do change D=R*T to R=D/T} and the last question is ,,, where do u get the Rate from..there is only distance and time... so plz help
No, you didn't break the law because Rule 124 of the Highway Code relates to motor vehicles and not to bicycles. Even if you went 120 mph it wouldn't have made a difference.
On only (N) x^3=[x(x+1)/2]^2 - [(x-1)x/2]^2. Defined the function f; f(z,x,y)=[z(z+1)/2]^2 - { [[x(x+1)/2]^2+[y(y+1)/2]^2 }. So f(z-1,x-1,y-1)=[z(z-1)/2]^2 - { [[x(x-1)/2]^2+[y(y-1)/2]^2 }. Impossible at same time both f(z,x,y) =0 and f(z-1,x-1,y-1)==0. If z^3=x^3+y^3 or z^n=x^n+y^n. ADIEU.
I'm really likely the grainy video circa 2007. The good old days.
"It was so long, by the way, nearly killed me".
heh heh ;)
I wish I was this happy about math....
If only all teachers were this enthusiastic.
@pyepyo1 This is just a rephrasing of the formula. If D=R*T and you want to solve for rate, you have to isolate R. The way you do that is to divide both sides by T, R*T/T = D/T, the T’s on the side with R divide out to be 1 which leaves you with R=D/T. If you plug in distance and time into this you can easily calculate the rate.
@MasterJake777 We've got a ton of them for free on our channel right now. You can also take a look at our blog and there are even more there!
I actually laughed more watching this video than i did watching dane cook's standup routine, and i learned some calculus. Thank you.
I freakin love you, no seriously I do. I'm taking Calc I next semester and just by watching your videos and looking over my textbook, I'm not so terrified anymore.
I love this guy's spirit :)
tbh. this is much really more clearer than me than the ton of pages in a book. plus your energy gave me more excitement.
after i watched your videos in precalculus and calculus you helped me a lot to understand what is going on in calculus books and what my prof is saying and i got (A) IN CALCULUS and im now good in physics thank you very much thinkwell and edward burger .... [ thinkwell = :) + :) ]
I've just really understood what calculus is about, thank you man!!!
Thank you! Your spirit makes it fun to watch these videos! can you make a video on time travel?
@curlyfryist Thanks! That is the perfect explanation!
Excellent!
@pyepyo1 Not sure what exactly you are asking here, but as long as you have distance and time, you can calculate rate from the D=R*T formula. Glad you enjoy our videos and hope we are able to help you jump ahead!
These videos actually did help!
ur videos help so much! thanks
AWESOME!
He stares into your soul at 8:44
Now when a Jr. high student understand the basics of calculus, you know this guy is a great teacher!
Yes, in this case the answer can be found with simple algebra. He used a very simple example. Calculus comes into play when you have a position function (for example I will use y=t^2. y being position and t being time.) Obviously, this has you changing speeds so it would be considerably harder to figure out some algebra to find his speed at say t=20 (if it is even possible at all). Calculus tells us he is going 40 miles / hour at t=20 if position is in miles. I hope you enjoy Calc it is
So 2000's, love it
@yondaime5685 Thank you! We love when people recommend our videos!!
nice job!
i love this guy...he's...metal \m/
Now i know how biologists calculated the area of an amoebae. Way to go prof.
Why didn't I have high school math teachers like this guy? I might have learned something and enjoyed it too.
The guy is a natural :)
@jkid1134 The Calculus in 20 minutes had to be fast since he was teaching so much in so little time.
An instant is an infinitely small portion of a second. However small a non-zero portion you come up with, an instant is smaller. So, nothing actually happens during an instant, because anything that takes time takes more time than an instant. However, because motion is continuous, the bike is still moving during an instant, even though it can't accomplish anything. So, the bike has a speed it travels during that instant, but the instant is so short that it covers zero distance in an instant.
the most interesting math class I've taken thus far. :)
you are my idol,,keep it up idol!!!you really connect!!keep it up!!your really funny and informative!!
@pyepyo1in simple terms if15= 5*3 then 5=15/3
Got dat AP test tomorrow
very beneficial
i like this guy =D
Hey i been tol i schmelll tooo? anyway..perhaps a Plank time measurement 10 to the minus 33rd power of magnitude? or the size of a single string as in string theorem..or most flavors of it anyway..
REALLLLY big tires, you can go any where on them actually
love the details that are irrelevant to calculus
Mrs. Jones I dont know where to write my post!
So I'm going to write it here.
Hi Mrs. Jones!
CLOSE THE BLAST DOORS!
@ThinkwellVids so you mean, you get the rate from the question ? correct?
i'm just a middle school student who want to jump a head a bit.so tyvm.. ur videos are awesome. :D
@curlyfryist awe.sweet. thank for more information.. help a lot XD
That was his average velocity. We're talking about instantaneous velocity. He could have been speeding up or slowing down during the bike ride, so the average velocity is not necessarily his instantaneous velocity at the 20 mile point.
Actually I should have said just limits because both derivatives and integrals are limits.
He rode at 20 mph.---5 miles in 15 minutes or 20 miles in 60 minutes.,or 90 minutes = 30 miles,9 minutes = 3 miles, 90x=30,9x=3,x=1/3, x*60=20
interesting
LMAO INSTRUCTING AND FUNNY =D
i don't get the part where you said { D,R,T ; D=R*T ; how do change D=R*T to R=D/T}
and the last question is ,,, where do u get the Rate from..there is only distance and time... so plz help
wishing you were my math teacher kind of hard to understant calculus at 15 =/
they should put a speed limit on how fast he talks, lol
he make coffee nerves
@GuyMcpersonLol Calculus is the most Graph heavy math ..........
thats what she said at 20 seconds
Yep, you basically use limits and derivatives to find both answers :)
Where are the answers to the two questions???
sounds like nardwuar if he was really really really smart lol. helpful information.
Is this like first grade math or something? Because this is what I did in school :o
Once you can find the area of a torus with algebra message me.
wow!
this comment makes my brain want to melt.
man i wish education was free
Gots fat tires it can go anywhere ! lol
wow lol
Lower voice. Increase microphone's pre-amp gain. Possibly lower caffeine intake.
*Torus
Where my Millis bros at
No vectors in a instantaneous measurement..and Heisenberg is not so sure bout dit..
I wish teachers had the mentality of teaching and not working through a long series of worksheets and tests
@TheRedAzuki First grade as in when you arelike 8
This video before 16 year can you imagine that 🙂 we are in 2024 now 🙂
a bike rider's speed and size are unrelated
N......
Heh, I took the Pre-Calculus course over the summer.
I'm in AP Calculus BC as junior. Looks good on a resume eh?
This seems slow now that I watched that other one.
So Ben Shapiro gave up on politics?
an ameoba..lol
No, you didn't break the law because Rule 124 of the Highway Code relates to motor vehicles and not to bicycles. Even if you went 120 mph it wouldn't have made a difference.
Arithmetic: Everything practical.
Algebra: How to graph it.
Geometry: Shapes.
Calculus: What he said.
way to eat the example
On only (N)
x^3=[x(x+1)/2]^2 - [(x-1)x/2]^2.
Defined the function f;
f(z,x,y)=[z(z+1)/2]^2 - { [[x(x+1)/2]^2+[y(y+1)/2]^2 }.
So
f(z-1,x-1,y-1)=[z(z-1)/2]^2 - { [[x(x-1)/2]^2+[y(y-1)/2]^2 }.
Impossible at same time both
f(z,x,y) =0 and f(z-1,x-1,y-1)==0.
If
z^3=x^3+y^3 or z^n=x^n+y^n.
ADIEU.
you are hyper 8O
Thumbs up if you are way to young for calculus.