Hey Patrick, your book that I ordered arrived in the mail today, I've never been so excited to get a maths book before. Frustratingly the exam of my final Calculus course is only weeks away and it was just recently that I realised you had a book published, luckily there's still some time to practice and this is exactly what I needed (good examples are so hard to find!!).
I hope you read this. I was wondering if you could do some more series tests examples. I've looked at all your old ones, but just was wondering if you could do an updated one. I have a final in 2weeks in calc2. ratio and root are easy. integral, telescoping, limit comparison I'm kinda getting lost
Bounded is very useful "word nuance" here implying there are intersection points otherwise you would need to take the definite integral of the difference of the absolute value of the integrand.
2:52 if you square both sides, aren't you supposed to put the condition that 6-x>=0 (==> 6>x), because x^(1/2) can't be negative, and that is why you should rule out the answer x=9 ???
+LIKAB055 I could help you out! Instead of thinking of the area beneath a polar curve as rectangles (as one would do with the rectangular co-ordinates), think of them as slices of a circle (the angle should be really small) with the centre at the origin. The equation for the area of a circle = (r^2*theta)/2 Since our angle is really small, it becomes (r^2 dtheta)/2 You can now integrate this from any to angles to get the area between that curve Really the secret here is to gain intuition about thinking of the slices with their centre at the origin Then when you have to find the area between 2 different curves. Look at this example: imgur.com/y6S7VIy The curve along the pi/2 line is 3sin(theta), and the curve along the 0 rad line is 3cos(theta) If you wanted to find the area between these 2, all you have to do is figure out the bounds on them. You can see the area between them starts at 0 rad and ends at pi/2 rad Imagine little slices that extend out from the centre to the functions. I'd even suggest taking out a paper and drawing them, if you find visualization tough. You will notice the area is bounded by 3cos(theta) from 0 to pi/4, and the area is bounded by 3sin(theta) from pi/4 to pi/2 So you just have to integrate (from 0 to pi/4) 1/2* [9cos(theta)^2 dtheta] and (from pi/4 to pi/2)1/2* [9sin(theta)^2 dtheta] Using a few trig identities ( cos^2(x) = (1+cos(2x))/2 and sin^2(x) = (1-cos(2x))/2 ), solving this becomes trivial :) Hope this helped, and I look forward to seeing Patrick's video on the same soon!
Multiply and divide left hand side with (sec A - tan A)....so it is =((sec A - tan A)/(sec A + tan A)(sec A - tan A)) =(sec A- tan A)/(sec^2 A - tan^2 A) =(sec A - tan A)/1..............since sec^2 A - tan^2 A is equal to 1 =1/cos A - sin A/ cos A =(1-sinA)/cosA
Come on! His explanations are too fast like hell. I know I could replay back and forth to catch up with his flow of explanation but I suppose he doesn't need to be in rush while explaining those who need help in Calculus.
too fast: someone bitches. too slow: someone bitches. middle pace: someone bitches. show too much: someone bitches. don't show enough: someone bitches. it be like that sometimes. or all the time on youtube. ask questions next time.
I didn't mean to criticize you but I just intended to share what I have experienced during the video. Please don't mind if I made you mad. Anyway, your lectures are pretty informative and beneficial and thanks a lot for that!
Hey Patrick, your book that I ordered arrived in the mail today, I've never been so excited to get a maths book before.
Frustratingly the exam of my final Calculus course is only weeks away and it was just recently that I realised you had a book published, luckily there's still some time to practice and this is exactly what I needed (good examples are so hard to find!!).
patrick, you're my savior!
After 8 years,your handwriting is awesome sir.Even with the left hand
I hope you read this. I was wondering if you could do some more series tests examples. I've looked at all your old ones, but just was wondering if you could do an updated one. I have a final in 2weeks in calc2. ratio and root are easy. integral, telescoping, limit comparison I'm kinda getting lost
Did you end up getting your degree?
@@ethanwhite1974 lol. Yes.
I have been searching for this, and now I finally found this.
This video was so helpful, thank you!!
Bounded is very useful "word nuance" here implying there are intersection points otherwise you would need to take the definite integral of the difference of the absolute value of the integrand.
Thank you, Patrick! This really helped me!
2:52 if you square both sides, aren't you supposed to put the condition that 6-x>=0 (==> 6>x), because x^(1/2) can't be negative, and that is why you should rule out the answer x=9 ???
yes that's right
Thank you so much for this! You make it look so easy!
u just saved my life
Great vid
thank you, very informative!
great explenation Patrick. I however used the quadratic formula since I suck when it comes to factoring lmao
Hi. How do you record your videos? What do you use to hold the camera like that?
Thanks a lot man
Please try to find for me area of a region bounded by a curve with root sign and a line
Please give the shadded portion of parabola and line
Do you have any video for area bound by 3 curves but one of the equation is just on x-axis . For eks x= 7
Why did you substitute X as 4 there
+Sunita Lata 4 is where the two non constant functions intersect, and where one integral stops and the other begins.
How can you solve for its volume? Like the same given and aroung x or y axis?
can you do a quick how to find area between 2 functions in polar
+LIKAB055 I could help you out!
Instead of thinking of the area beneath a polar curve as rectangles (as one would do with the rectangular co-ordinates), think of them as slices of a circle (the angle should be really small) with the centre at the origin.
The equation for the area of a circle = (r^2*theta)/2
Since our angle is really small, it becomes (r^2 dtheta)/2
You can now integrate this from any to angles to get the area between that curve
Really the secret here is to gain intuition about thinking of the slices with their centre at the origin
Then when you have to find the area between 2 different curves.
Look at this example: imgur.com/y6S7VIy
The curve along the pi/2 line is 3sin(theta), and the curve along the 0 rad line is 3cos(theta)
If you wanted to find the area between these 2, all you have to do is figure out the bounds on them. You can see the area between them starts at 0 rad and ends at pi/2 rad
Imagine little slices that extend out from the centre to the functions. I'd even suggest taking out a paper and drawing them, if you find visualization tough.
You will notice the area is bounded by 3cos(theta) from 0 to pi/4, and the area is bounded by 3sin(theta) from pi/4 to pi/2
So you just have to integrate
(from 0 to pi/4) 1/2* [9cos(theta)^2 dtheta]
and
(from pi/4 to pi/2)1/2* [9sin(theta)^2 dtheta]
Using a few trig identities ( cos^2(x) = (1+cos(2x))/2 and sin^2(x) = (1-cos(2x))/2 ), solving this becomes trivial :)
Hope this helped, and I look forward to seeing Patrick's video on the same soon!
from where are You?
where are you from*
can you help me with inverse trigonometric function?
cant u integrate just 1 to 4 in te integral of [[ Sqrt(x) + 1 ] - [7-X]
no
This has nothing to do with the vid but how do i prove 1/(secθ + tanθ) = (1-sinθ)/cosθ
Multiply and divide left hand side with (sec A - tan A)....so it is
=((sec A - tan A)/(sec A + tan A)(sec A - tan A))
=(sec A- tan A)/(sec^2 A - tan^2 A)
=(sec A - tan A)/1..............since sec^2 A - tan^2 A is equal to 1
=1/cos A - sin A/ cos A
=(1-sinA)/cosA
You can prove it like that....or go from the right hand side
Multiply and divide the right hand side with (Sec A + tan A)
Ted -Ed
Is there anyone?i need a math solution very badly.please answer me if anyone want to help me
Come on! His explanations are too fast like hell. I know I could replay back and forth to catch up with his flow of explanation but I suppose he doesn't need to be in rush while explaining those who need help in Calculus.
too fast: someone bitches. too slow: someone bitches. middle pace: someone bitches. show too much: someone bitches. don't show enough: someone bitches.
it be like that sometimes.
or all the time on youtube.
ask questions next time.
I didn't mean to criticize you but I just intended to share what I have experienced during the video. Please don't mind if I made you mad. Anyway, your lectures are pretty informative and beneficial and thanks a lot for that!
@@patrickjmt we need a formula for optimum speed
this is poggers