For part D if you want to solve it easier, take the x value (being pi/2) and create a number line. Then you plug in a number less than pi/2 (0) and greater than pi/2 (pi) and you would see that there would be a positive to negative change therefore being a maximum.
@@dydwnsTJ These type of questions are asked very often on ap exams, or just used a lot in calculus classes, so it may be helpful to somebody to know that when they see this sort of set-up they can do a sign chart.
In Part c , I think the only point to be considered was ( pi/2,2) since y had to be positive. That explains why part d is asking whether the point obtained in c is a max or minimum. Finding the second derivative and plugging (pi/2 ,2) is negative meaning concave down and signifying a maximum point. Great job overall.
This is a description of what I did in the video. Some other commenters have suggested an alternative method which I think is probably equally good and less work.
for 5b, i left my answer in the sqrt(3)cos(0)/4sqrt(3)-sin(0) because i honestly didn't know cos/sin of 0 and then i put that into the tangent equation. is this fine?
For part D if you want to solve it easier, take the x value (being pi/2) and create a number line. Then you plug in a number less than pi/2 (0) and greater than pi/2 (pi) and you would see that there would be a positive to negative change therefore being a maximum.
Bro who cares It's all over
I agree, sign analysis to determine positive/negative on either side of the zero.
@@dydwnsTJ for the people that are doing calculus III and still have to do something like this.
@@dydwnsTJ These type of questions are asked very often on ap exams, or just used a lot in calculus classes, so it may be helpful to somebody to know that when they see this sort of set-up they can do a sign chart.
What equation are u plugging in 0 and pi into?
In Part c , I think the only point to be considered was ( pi/2,2) since y had to be positive. That explains why part d is asking whether the point obtained in c is a max or minimum. Finding the second derivative and plugging (pi/2 ,2) is negative meaning concave down and signifying a maximum point. Great job overall.
This is a description of what I did in the video. Some other commenters have suggested an alternative method which I think is probably equally good and less work.
for 5b, i left my answer in the sqrt(3)cos(0)/4sqrt(3)-sin(0) because i honestly didn't know cos/sin of 0 and then i put that into the tangent equation. is this fine?
Yes! You don't need to fully simplify your answer if you feel you'll make a mistake.
I feel so good about this frq
Fr bro