For 6d, I used dy/dt = dy/dx * dx/dt to arrive at the same solution; sub in (1/2, -2) for x and y respectively in the dy/dx equation to find that part, then subbed in 2/3 for dx/dt; a lot less messy than doing imp diff with product rule again
for the last one you only need to find dy/dx at that point which is -4/3 and multiply it by dx/dt which is 2/3 and the dx cancel out giving dy/dt which is -8/9
for 6b and 6c, I got the right answers but for horizontal I accidently solved for vertical and vice versa. would I get no points or just like partical credit for a mistake?
I'm almost positive no points. When they grade by guidelines they look at single question individually for specific equations/explanations from what I understand.
there's a chance you get some points depending on what kind of work you did. we'll never know, but hopefully you've got enough points on the exam to get the score you want!
nope, it's a great method. better than what i did, but at least my approach works. (i wonder what the split is on people using one vs. the other approach.)
yeah, that's a great solution. it didn't even occur to me as i was solving this. (i guess in AB I tend to not think of that option, while on BC i would think it's the most obvious way to deal with parametrics...context matters in my brain!)
@@turksvids I know what you mean - what you did in this video is generally how I would attack this problem in AB, but not in BC because of the context in parametric equations or polar coordinates.
For 6d, I used dy/dt = dy/dx * dx/dt to arrive at the same solution; sub in (1/2, -2) for x and y respectively in the dy/dx equation to find that part, then subbed in 2/3 for dx/dt; a lot less messy than doing imp diff with product rule again
yeah, it's a good move. didn't occur to me at the time.
for the last one you only need to find dy/dx at that point which is -4/3 and multiply it by dx/dt which is 2/3 and the dx cancel out giving dy/dt which is -8/9
good solve.
for d would we have to put units or not? for example dy/dt= -8/9 units per second
I solved for x on the exam and did some nasty ass calculations 💀
yeah, it was an option...i wanted no part of that square root that was going to show up.
for 6b and 6c, I got the right answers but for horizontal I accidently solved for vertical and vice versa. would I get no points or just like partical credit for a mistake?
I'm almost positive no points. When they grade by guidelines they look at single question individually for specific equations/explanations from what I understand.
Ooo, not sure. AP readers would probably discuss that situation. Hopefully they grade in your favor.
there's a chance you get some points depending on what kind of work you did. we'll never know, but hopefully you've got enough points on the exam to get the score you want!
Why not just multiply dy/dx by dx/dt on part d and evaluate that on the given point? Am I missing something?
nope, it's a great method. better than what i did, but at least my approach works. (i wonder what the split is on people using one vs. the other approach.)
did we need units for the last part i got it right but didn’t put units
only question i missed on all frq was the vertical tangent 😭 i set up the equation just solved it wrong
yea I think u did. I FOROGT UNITS ALSO LMFAO. its j 1 point off
No, only if they asked you to indicate units.
@user87590 nah, if they don't mention that you need them, you don't need them.
@user87590 awesome work, though!
For 6d, could you also say that (dy/dt) = (dy/dx)*(dx/dt) to avoid redoing the differentiation?
yeah, that's a great solution. it didn't even occur to me as i was solving this. (i guess in AB I tend to not think of that option, while on BC i would think it's the most obvious way to deal with parametrics...context matters in my brain!)
@@turksvids I know what you mean - what you did in this video is generally how I would attack this problem in AB, but not in BC because of the context in parametric equations or polar coordinates.
zero cubbed not squared
Heh