If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 11

Lesson 3: AP Calculus BC 2011

# 2011 Calculus BC free response #1a

Velocity and acceleration vectors for particle. Created by Sal Khan.

## Want to join the conversation?

• In this problem you evaluate "cos 9" with the calculator. For the AP test, are all trig values assumed to be in radians or degrees, or was it specified somewhere not shown in problem? Obviously cosine of 9 degrees is very different than cosine of 9 radians.
• When you're simply given a number like this 99.9% of the time you're working with radians, especially if they want you to differentiate or integrate.

If they write something like theta = x degrees or there's a degrees sign you'll know you're working with degrees.
• are polar coordinates, parametric equations, and conics on the Calculus BC AP exam?
• Yes, on the BC exam, no for the AB. Usually only a small portion though.
• What if I showed this as my work, instead of Sal's engineering-style vectors:

Speed = sqrt( x' ^2 + y' ^2)

Would that be okay?
• Yes. If you directly applied the formula Speed = sqrt( x' ^2 + y' ^2), you would get the same result Sal derived in the video.
(1 vote)
• The answer to the acceleration vector problem would actually be <4, -5.467>, because it's asking for a vector, not a definite solution (like the speed or distance formula). That was simply found by taking the 2nd Derivative of x and y: <x''(3), y''(3)>
• <4, -5.467> and 4 i + -5.467 j are 2 equally valid ways of expressing an vector because they clearly denote the x and y component of the vector.