Thank you a million times. Imagine I could not figure this out the whole week, but luckily for me , I had to turn to my best maths clips ever since - "Patrick JMT videos on you tube." I cant emphasize enough on the clarity of your explanations. I wish you knew how gifted you are. Once again, am so greatful for your work. Thanks.
I wish I could give you money instead of the school for my Calc 2 class at my college because you're basically the reason I have a good grade, let alone a passing one. Thanks so much!!!
I know people leave comments thanking you all the time but I still don't think you understand how appreciative we are. You are the only reason I made it this far in math. Saving me a lot of stress and money. Thanks.
I just saw this in class and thanks to your video, which explains better than my teacher, I finally understand the process: especially for the y=mx+b form of the answer, I didnt know it was unnecessary.. You have my gratitude! 5/5! I will keep on watching your videos Maestro!
All of your videos are great, I consult them before all of my Cal 2 tests and they help a ton. For once watching youtube before a test is actually a good thing.
Calc II is pretty hard, but you're easing the pain quite a bit with these videos! I've got a test today, and I tried studying my notes, but I didn't really understand parametric equations. Then I saw this video, and it all makes sense!
parametric function is when the x and y coordinates are each functions of another variable, usually denoted as 't' i should have a video on it somewhere : )
Hmm...these videos will be useful when I'm revising before a test. I do 3 Unit Maths in Australia, which is basically the second hardest level. Thanks a lot.
Nice video, dude. I still don't even know what a parametric function is, but I saw it in my textbook and in this video and I think I can do this now! I just don't know exactly what I'm calculating yet. I haven't taken calculus yet! Lets do this!
Wow, I can't believe I didn't watch these videos before taking my math final today. I failed unfortunately, but next semester these videos will come in handy.
Hey Patrick how would I find the slope of the perpendicular line to y=f(t) at t=2 if I'm given y=2x/(x+1) and t=x^3+x? I figure this is a parametric problem but I'm used to getting both x and y equations with respect to t.
What I've I'm given two parametric equations, a point that the tangent line passes through (that I don't think is even on the curve) and I'm supposed to find the equation of the tangent line? When I calculate my t values, they don't match for the point that the tangent line is supposed to pass through.
I'm so glad I found your videos! Do you have any videos similar to this but also finding all horizontal tangent lines? I know this means the slope is zero but I can't find the second point!
What do you do if dx/dt at the given value for t IS equal to zero? I'm given the point (1,0) and when you solve to find common value you find t=0. x=t^2+1, so dx/dt=2t. Substitute in t=0 and I end up with the slope -1/0. (y=t^3-t dy/dy=3t^2-1). Is this a legitimate answer or is there another step?
Hey patrick, is there any way to find a general derivative to parametric equations rather than only finding the derivatives. This is purely out of curiosity.
No, the root of 4 is only +2. But the root of x^2 is +/-x. This is because x is an unknown and thus it may have different values. But by itself, the square root of a known number is always the positive one. Hope that helps! :)
Please, I need some help! Here is the question: Let P be a point on the curve x(t^2), y=(1/t). If the tangent to the curve at P meets the x- and y-axes at A and B respectively, prove the PA=2PB. I have found the gradient of the tangent at point P (through differentiation of x and y) and I have found the equation of the tangent. I have also found points A and B by using the equation of the tangent. I got: A (3(t^2), 0) and B (0, (3/2t)). Now, I am stuck. What do I do? Please show the steps! Thank you!!!
I hope you're making millions, Patrick.
ha, far far far far far from it
But you're worth a million in my eyes. Thanks Pat
in that case, head on over to 'Patreon' and make a donation! ;)
that last part never seems to work... why pay for the cow when you get the milk for free! :)
lol. I'm sorry i'm a broke college kid. One day hopefully i'll be an engineer and Pat will get all the cows :)
Thank you a million times. Imagine I could not figure this out the whole week, but luckily for me , I had to turn to my best maths clips ever since - "Patrick JMT videos on you tube." I cant emphasize enough on the clarity of your explanations. I wish you knew how gifted you are. Once again, am so greatful for your work. Thanks.
I wish I could give you money instead of the school for my Calc 2 class at my college because you're basically the reason I have a good grade, let alone a passing one. Thanks so much!!!
I know people leave comments thanking you all the time but I still don't think you understand how appreciative we are. You are the only reason I made it this far in math. Saving me a lot of stress and money. Thanks.
It's 2017 and I'm still watching you videos. Thank you SO much. This really helps me a ton and you are saving my calc grade rn haha.
2018 for me and they are amazing
I've read several calculus "help books" and three textbooks, and your videos make more sense than any of them.
Over 10 years later and we're still coming to you for help. Thank you so much!!
my pleasure :)
I just saw this in class and thanks to your video, which explains better than my teacher, I finally understand the process: especially for the y=mx+b form of the answer, I didnt know it was unnecessary..
You have my gratitude! 5/5! I will keep on watching your videos Maestro!
All of your videos are great, I consult them before all of my Cal 2 tests and they help a ton. For once watching youtube before a test is actually a good thing.
Best math channel ever! Thanks Patrick from Albania :D
Calc II is pretty hard, but you're easing the pain quite a bit with these videos! I've got a test today, and I tried studying my notes, but I didn't really understand parametric equations. Then I saw this video, and it all makes sense!
parametric function is when the x and y coordinates are each functions of another variable, usually denoted as 't'
i should have a video on it somewhere : )
by definition, the square root of a positive number is positive.
so, sqrt(25) = 5
sqrt (100) = 10
im always impressed with how easy you make this stuff. thankyou sooooo much. you are getting me through calc 2.
Holy crap!! That Total Recall movie preview you have as your advertisement looks amazing!
no problem!
long live PatrickJMT!
@poxy435 thanks for the very kind words :)
You're a great teacher! Thanks a lot!
Hey Patrick, do you have a video on the direction the parametric graphs go?
Thanks Sir. Your hit the nail on the head. Now I am pretty sure to tickle any parametric equations by using your example!
Thumps up sir!
God bless
Man i am acing calculus becoz of you, i was really bad at maths, thanks a lot !
i like it better when you use the long sheet of paper as opposed to the dry erase board. thanks for the great vids!
Hmm...these videos will be useful when I'm revising before a test. I do 3 Unit Maths in Australia, which is basically the second hardest level. Thanks a lot.
Man you really Are good at teaching lol. I've been watching your vids like crazy for calculus concept overviews. xD
Nice video, dude. I still don't even know what a parametric function is, but I saw it in my textbook and in this video and I think I can do this now! I just don't know exactly what I'm calculating yet. I haven't taken calculus yet!
Lets do this!
Fantastic, as always. When are you coming out with your book?
Once again, total life saver, thanks Patrick!
i agree
ya!
you're really easy to follow...... thanks for posting
Wow, I can't believe I didn't watch these videos before taking my math final today. I failed unfortunately, but next semester these videos will come in handy.
@patrickJMT You should change most of them to instream ads..those have a better ctr from what i've noticed.
why cant we plug-in the y and x values for x and y instead od x1 and y1?
Hey Patrick how would I find the slope of the perpendicular line to y=f(t) at t=2 if I'm given y=2x/(x+1) and t=x^3+x?
I figure this is a parametric problem but I'm used to getting both x and y equations with respect to t.
What I've I'm given two parametric equations, a point that the tangent line passes through (that I don't think is even on the curve) and I'm supposed to find the equation of the tangent line? When I calculate my t values, they don't match for the point that the tangent line is supposed to pass through.
I'm so glad I found your videos! Do you have any videos similar to this but also finding all horizontal tangent lines? I know this means the slope is zero but I can't find the second point!
i agree, a good vid is nice!
books are always good though!
Thank you - you are an excellent video maker and math teacher and you saved my ass so many different times
I would just sincerely like to thank you :)
What do you do if dx/dt at the given value for t IS equal to zero? I'm given the point (1,0) and when you solve to find common value you find t=0. x=t^2+1, so dx/dt=2t. Substitute in t=0 and I end up with the slope -1/0. (y=t^3-t dy/dy=3t^2-1). Is this a legitimate answer or is there another step?
Thanks, Patrick!!! You help me a lot!
AWESOME. Very clear explanation :)
patrick, i would be completely hopeless in calculus 2 if it wasn't for you. i feel very lucky that we live in the youtube age.
who's the man?!
: )
YOU THE MAN!!!!!!!!!
What if the x value when finding the slope is zero?
hi, I have a question: What happens if after differentiate dy/dx and it contains t?
Hey patrick, is there any way to find a general derivative to parametric equations rather than only finding the derivatives. This is purely out of curiosity.
ha! i did not know i had a book coming out!
I just want to say thank you
shoutout to you for helping me get an A in calc 2 :)
You're the man bro, thanks a lot!
No, the root of 4 is only +2. But the root of x^2 is +/-x. This is because x is an unknown and thus it may have different values. But by itself, the square root of a known number is always the positive one. Hope that helps! :)
dude ur the best!
awesome teaching man .keep it up and help me pass calculas 2 with an A
@GeekBoy03 yeah but you do have to consider that he is making these videos for free.
I wish my math teacher would watch your videos
thanks friend : )
great, but i thought you'd have to do -1/m to get the gradient of the tangent?
You got great handwriting!
Thanks PatrickJMT!!!!
thank you sir, cleared that up for me!
@GeekBoy03 google posts ads on all videos, if you have not noticed. and i will agree with the earlier person: stop whining.
even though this question was 6 months ago, for future reference, that means that the slope is undefined at that point, ie. there is no derivative.
Very helpful. Thank you.
Shouldn't the slope be 23/4?
Imagine what would happen to us if it weren't for your videos... it boggles me how people go though school back in the 90s when there was no "TH-cam"
Please, I need some help!
Here is the question: Let P be a point on the curve x(t^2), y=(1/t). If the tangent to the curve at P meets the x- and y-axes at A and B respectively, prove the PA=2PB.
I have found the gradient of the tangent at point P (through differentiation of x and y) and I have found the equation of the tangent. I have also found points A and B by using the equation of the tangent. I got: A (3(t^2), 0) and B (0, (3/2t)).
Now, I am stuck. What do I do? Please show the steps! Thank you!!!
you saved the day!
thanks vids helped a lot
Couldn't you also eliminate the parameter to turn the parametric equation into a function, and then take the derivative then?
thanks helped a lot!
word
that would find the line perpendicular to the tangent going through the point.
you are a saint
what if the t is not given
tks sir....
i'd like to see the proof for that equation please
Nice handwriting.
@GeekBoy03 Why complain, still a great video.
A slope of -1/0?
Thank you so much!
I'm scared of the time when I reach a level in math which Patrick doesn't cover in his videos anymore
THANK YOU!
awesome, thankyou so much!
Make babies Pat, the world needs 'em. NOW.
TY
Thanks a lot.
Thank you!!!
Who needs a books, videos are better. We all have textbooks anyway.
thank you!
If this is basic 1st year calculus. i cant wait for 4th year..... if i get there
thanks dude
Thank u sir
Algebra II + Calculus = useful
but the questions is, WHAT IF X AND Y ARE FUNCTIONS OF TWO PARAMETERS. WAT DO THEN????
what about at a point
Again, Sorta, again, kinda good, again
algebra is the hardest part of calculus
Thanks! :D
I wish you were my calc professor haha