For anyone wondering in the future, the horizontal tangent line of (0, pi/2) does not exist. However, (0, pi/2) does exist as a vertical tangent line. This is because when you plug in a "pi/2" into the derived equation, you get 0/0. Whenever we see a derivative with the result 0/0, we can use L'Hopital's rule to find the true value of the equation. After using L'Hopital's rule, we get 1/0 when we plug in pi/2, meaning that the tangent line (0, pi/2) is vertical, as the zero is in the denominator. IF, on the other hand, the zero was in the numerator, the tangent line at (0, pi/2) would be horizontal. Hope that helped. :)
I missed some notes for some reason in class and I couldn't figure out on my own why the (0, pi/2) points weren't included in the answer. This video made everything clear, and now I know to check the points on my test. It was such an easy solution to my puzzlement! Thank you.
Thank you for a superb explanation. Your videos need more exposure. I always tell other students to look for your videos. I am grateful to individuals like yourself who take the time to help others struggling with the concepts in math.
+Mathispower4u !! Why can we write a vertical tangent on the sharp point 10:31?? You just said we couldn't draw the tangent there 5:43! My Webassign agrees with you!
Very helpful video. However my textbook says that when you get the same point for both horizontal and tangent line (like the (0,pi/2) in this problem), you have to take the limit of the derivative using L'Hopital's rule. I also wonder if you can skip that and just see if the sharp point is facing left or right and conclude that the tangent is horizontal or if the sharp point is facing up or down the tangent is vertical.
Briahna, you are correct. But we are not looking at the inverse sine of 1/2. Instead, we are trying to find out where the sine is equal to 1/2. Seems like there is no distinction, but there is - sine actually equals 1/2 for an infinite number of angles! The inverse sine only gives reference angles. You need to know that sine equals 1/2 in two different quadrants, the first and second. The inverse sine of 1/2 is pi/6, so this is the reference angle in those two quadrants.
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For anyone wondering in the future, the horizontal tangent line of (0, pi/2) does not exist. However, (0, pi/2) does exist as a vertical tangent line. This is because when you plug in a "pi/2" into the derived equation, you get 0/0. Whenever we see a derivative with the result 0/0, we can use L'Hopital's rule to find the true value of the equation. After using L'Hopital's rule, we get 1/0 when we plug in pi/2, meaning that the tangent line (0, pi/2) is vertical, as the zero is in the denominator. IF, on the other hand, the zero was in the numerator, the tangent line at (0, pi/2) would be horizontal.
Hope that helped. :)
I missed some notes for some reason in class and I couldn't figure out on my own why the (0, pi/2) points weren't included in the answer. This video made everything clear, and now I know to check the points on my test. It was such an easy solution to my puzzlement! Thank you.
This is a bitchload amount of work for one problem
Thank you for a superb explanation. Your videos need more exposure. I always tell other students to look for your videos. I am grateful to individuals like yourself who take the time to help others struggling with the concepts in math.
Your demonstrations are very much appreciated.
+Mathispower4u !! Why can we write a vertical tangent on the sharp point 10:31?? You just said we couldn't draw the tangent there 5:43! My Webassign agrees with you!
Shouldn't there be a vertical tangent since point (0,pi/2) is undefined?
very helpful. better than my textbook's video explanations
@maximus5415 I'm getting an error trying to explain why it is correct. I'll send you a message.
Very helpful video. However my textbook says that when you get the same point for both horizontal and tangent line (like the (0,pi/2) in this problem), you have to take the limit of the derivative using L'Hopital's rule. I also wonder if you can skip that and just see if the sharp point is facing left or right and conclude that the tangent is horizontal or if the sharp point is facing up or down the tangent is vertical.
very clear and helpful thank you so much
buena explicación saludos desde Colombia
Im confused I thought sin inverse had a range of (-pi/2,pi/2)
Briahna, you are correct. But we are not looking at the inverse sine of 1/2. Instead, we are trying to find out where the sine is equal to 1/2. Seems like there is no distinction, but there is - sine actually equals 1/2 for an infinite number of angles!
The inverse sine only gives reference angles. You need to know that sine equals 1/2 in two different quadrants, the first and second. The inverse sine of 1/2 is pi/6, so this is the reference angle in those two quadrants.
very helpful, I really appreciate.. the only thing I can't understand was the quote at the end of the video 😂
Thank you!
great videos
thank you
Thank you, it helped a lot ;)
Bless up though.
That's what I'm talking about.
THANK YOUUUUUU
thx
love you, lol.
hi
Thank you!