Calling all smart people - physics help!

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
atalkinghead

atalkinghead

Full Member
10+ Year Member
15+ Year Member
Joined
Jun 23, 2007
Messages
166
Reaction score
0
Points
0
Location
Albuquerque
  1. Pre-Medical
Members do not see this ad.
Hey everyone ... so I am trying to study physics (I haven't taken a physics course in 2 years!!!).

I was hoping someone could PLEASE help me.

Okay ... I am going to sound incredibly stupid, but ....

in the first chapter in the kaplan book for "Units and kinematics" they calculate the following:

D^2=900 + 2500
D^2=3,400 (I can get this far, hehe)
D=10 sqrt(34)

HOW do you estimate the sqrt of 3400? 😕

Also they took arctan(1.666)=59

HOW do you estimate the arctan? 😱 😕

Thanks for reading this and I REALLY appreciate all your help. 😀😀
 
Sort by date Sort by votes
for sqrt of 3400 I would think of it as 3600 and than to 36 which has a sqrt of 6. so from there i would lower just a little bit to about 5.7 to 5.8. So than it would be around 57-58.

arctan- no clue, memorize 0,30,60,90,120etc

hope that helped on the sqrt
 
Upvote 0 Downvote
sqrt of 3400 is easy. this is what I do:

I usually try to break apart a number into numbers with whole numbers as square roots. for example, sqrt3400 = sqrt(34*100) = sqrt34 x sqrt100 = ~6 x 10 = ~60. you should remember that for the mcat such estimations are good enuf.

as for the arctan, i doubt that the mcat would expect us to know that.

-minhaj
 
Upvote 0 Downvote
For square roots, just approximate it - it will be more than enough on the MCAT. As for the arctan or other trigonometry calculations, knowing them is not necessary. Values will be given in the passage like sin(60)= whatever and cos(45)= whatever.
 
Upvote 0 Downvote
atalkinghead said:
D^2=900 + 2500
D^2=3,400 (I can get this far, hehe)
D=10 sqrt(34)

HOW do you estimate the sqrt of 3400? 😕
Click to expand...

I'm not sure where you got lost (you remarked on your ability to get to the second step), but are you questioning how they arrived at the last part:

D=10 sqrt(34)

??
If you are, I can help out a lil bit--

sqrt(3400) = sqrt(34*100)
& you know sqrt(100) = 10
So, "10" pops out from underneath the square root sign.
Now you have: (10)sqrt(34)
& as mentioned by others, you can estimate sqrt(34) to be 5 and 6, but much closer to 6. Let's just say it's 5.8
Therefore, your final answer would be (10*5.8) = 58
 
Upvote 0 Downvote
If you need better precision, and the function is fairly simple, just run a quick taylor series in your head - it's just an extra term to the "3400 is close to 3600" method mentioned above:

f(x) ~ f(x0) + f'(x0)*(x-x0)

f(x) = x^(1/2)
f'(x) = (1/2)x^(-1/2)

3600 is near by and a perfect square, so

f(3600) = 60
f'(3600) = (1/2)*(1/60) = 1/120
(x-x0) = -200

f(3400) ~ 60 - 200/120 ~ 58.3
 
Upvote 0 Downvote
pyromatic said:
If you need better precision, and the function is fairly simple, just run a quick taylor series in your head - it's just an extra term to the "3400 is close to 3600" method mentioned above:

f(x) ~ f(x0) + f'(x0)*(x-x0)

f(x) = x^(1/2)
f'(x) = (1/2)x^(-1/2)

3600 is near by and a perfect square, so

f(3600) = 60
f'(3600) = (1/2)*(1/60) = 1/120
(x-x0) = -200

f(3400) ~ 60 - 200/120 ~ 58.3
Click to expand...

Haha, just remember that neither Taylor nor McLauren ever took the MCAT. They would have been too slow. 😉
 
Upvote 0 Downvote
The trig you'll have to know for the mcat are sin/cos of 0, 30, 45, 60, 90 degrees.

sin 0 = sqrt(0)/2
sin 30 = sqrt(1)/2
sin 45 = sqrt(2)/2
sin 60 = sqrt(3)/2
sin 90 = sqrt(4)/2

cos 0 = sqrt(4)/2
cos 30 = sqrt(3)/2
cos 45 = sqrt(2)/2
cos 60 = sqrt(1)/2
cos 90 = sqrt(0)/2
 
Upvote 0 Downvote
sqrt (2500) < sqrt (3500) < sqrt (3600)

50 < sqrt (3500) < 60

Since sqrt (3500) is closer to sqrt (3600), it can be approximated to be ~58/59.
 
Upvote 0 Downvote
atalkinghead said:
Also they took arctan(1.666)=59

HOW do you estimate the arctan? 😱 😕
Click to expand...

Yeah, I hated it when Kaplan just breezily took arctans and inverse sines and stuff like that without any explanation. Doing things like that are beyond the scope of the MCAT, so you won't have to do that stuff. You will need to know sine/cosine/tangent of basic angles like 0/30/45/60/90/180 but I think that's about it.

I think Kaplan goes above and beyond so you see the process of arriving at the exact correct answer even though the MCAT only requires you to estimate so you can eliminate answer options.
 
Upvote 0 Downvote
Thank you all so much! I get it now. 😀

I have become so dependent on my calculator that I was confused on how to estimate.

I also got confused on how they went from sqrt of 3400 to 10*sqrt of 34, but I get it now thanks to you guys.

👍

I am so grateful!!!!
 
Upvote 0 Downvote
MrMattOglesby said:
I'm not sure where you got lost (you remarked on your ability to get to the second step), but are you questioning how they arrived at the last part:

D=10 sqrt(34)

??
If you are, I can help out a lil bit--

sqrt(3400) = sqrt(34*100)
& you know sqrt(100) = 10
So, "10" pops out from underneath the square root sign.
Now you have: (10)sqrt(34)
& as mentioned by others, you can estimate sqrt(34) to be 5 and 6, but much closer to 6. Let's just say it's 5.8
Therefore, your final answer would be (10*5.8) = 58
Click to expand...

Mr. Matt ... that's exactly where I was confused. Thanks 🙂
 
Upvote 0 Downvote
Hello all.

I am new to the forum and I am having a similar problem. I did find a webpage that says that we will definitely experience the arcsin, arccos, and you guessed it, the acrtan! Come and see the webpage here for yourself.
http://www.aamc.org/students/mcat/preparing/psprep.htm

I really need to know how you can estimate the arctan on the MCAT. They pull stuff out in the Kaplan book like. Arctan(3) = 72 degrees and arctan(4/3) = 53 degrees. Other than memorize the "pythagorean triple"
http://www.mathopenref.com/triangle345.html
and other common triangles I don't think I can find an easy way to get through these problems that say "Oh and find the angle too; Tan(theta) = 4/3, find the angle". I could sure use some direction here.
 
Upvote 0 Downvote
You must log in or register to reply here.

Similar threads

P
Physics Question for Next Step/Blue Print Question Bank
Replies
5
Views
12K
doctadank
doctadank
T
  • Question Question
NS Physics Question
Replies
1
Views
4K
RunningMSN
R
treeclimbingmonkey
  • Question Question
OG Spoiler Question Physics
Replies
3
Views
2K
FrameshiftAndrew
FrameshiftAndrew
Mkif586
  • Question Question
Bad at physics...
Replies
6
Views
3K
zorcandpals
zorcandpals
H
  • Question Question
Kaplan MCAT Physics Question - unlimited velocity
Replies
2
Views
2K
Hyein
H
Top Bottom