AP®︎/College Calculus BC
- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
2015 AP Calculus BC 2a
x coordinate of particle at a certain time.
Want to join the conversation?
- why did Sal have to add the integral of cos(x^2) in order to solve part a?(2 votes)
- That is the velocity function of the x coordinate of the particle. We have the initial x position as 3, so we add the integral of the x coordinate velocity function with bounds 1 and 2, which gives us the distance the x coordinate has moved from t = 1 to t = 2.(1 vote)
- Is possible to solve these types of integrals without a calculator?(1 vote)
- If you mean the integral of cos(t^2), no. It cannot be calculated by hand.(1 vote)
- So I understand the lesson, but tried doing it another way initially by integrating cos(t^2) and solving for c before plugging back t=2. However, I got a different answer (2.39006) when doing this. Am I doing something wrong, or does this method not work?(1 vote)
- Are these problems calculator or non-calculator?(0 votes)
- Problems 1 and 2 on the Calculus AB and BC test are calculator, and 3-6 are non-calculator.(4 votes)
- Would it be necessary to put my TI into parametric mode as opposed to functions when doing these parametric/calculus questions?(1 vote)
- Your calculator does not have to be in parametric mode for what Sal was doing (using the solver). This is because Sal put the parametric equation in terms of X (instead of t), and was using the traditional f(x)=(x,y) function notation instead of f(t)=(x(t),y(t)).
However, if you wanted to graph the parametric equation, you would need to put your calculator in parametric mode. For the most part, moving your calculator from "Func" to "Par" to "Pol" to "Seq" is for graphing, not for solving equations. If you put your calculator in Par (parametric mode), it would not affect the outcome, but it would change the variable to "t".(0 votes)
- Wow, thanks for the help, i am totally going to use a calculator on my test. Jut kidding, but thanks for he help TEEHEE(0 votes)
- You do realize you are allowed to use a graphing calculator for the first 2 problems of the FRQs and the last 15 problems of the MCQs, right?(1 vote)
- [Voiceover] "At time t is greater than or equal to zero, "a particle moving along a curve in the xy-plane "has position x of t and y of t." So its coordinate is given by the parametric function x of t, and y-coordinate by the parametric function, y of t. "With velocity vector v of t is equal to," and the x component of the velocity vector is cosine of t squared, and the y component of the velocity vector is e to the 0.5t. "At t equals one, "the particle is at the point 3, comma, 5." Is at the point 3, comma, five. All right. "Find the x-coordinate of the position of the particle "at time t is equal to two." All right, so how do we think about this? Well, you could view the x-coordinate at time t equals two. So let's say, so we could say x at time two, which they don't give that to us directly, but we could say that's going to be x of one plus some change in x as we go from t equals one to t equals two. But what is this going to be? Well, we know what the velocity is. And so the velocity, especially the x component, and we can really focus on the x component for this first part 'cause we only wanna know the x-coordinate of the position of the particle. Well, we know the x component of velocity as a function of t is cosine of t squared. And if you take your velocity in a certain dimension and then multiply it times a very small change in time, in a very small change in time, dt, this would give you your very small change in x. If you multiply velocity times change in time, it'll give you a displacement. But what we can do is we can sum up all of the changes in time from t equals one to t equals two. Remember, this is the change in x from t equals one to t is equal to two. So what we have right over here, we can say that x of two, which is what we're trying to solve, is going to be x of one, and they give that at time equals one, the particle is at the point three, comma, five. Its x-coordinate is three. So this right over here is three. And then our change in x from t equals one to t equals two is going to be this integral. The integral from t equals one to t equal two of cosine of t squared, dt. And just to make sure we understand what's going on here, remember, how much are we moving over a very small dt? Well, you take your velocity in that dimension times dt, it'll give you a displacement in that dimension, and then we sum 'em all up from t equals one to t equals two. And in this part of the AP test, we're allowed to use calculators and so let's use one. All right. So there's my calculator. And I can evaluate. So let's see, I wanna evaluate three plus the definite integral. I click on Math, and then I can scroll down to function integral, right there. The definite integral of, and I make sure I'm in radian mode, that's what you should assume, unless they tell you otherwise. Cosine of t squared, now I'll use x as my variable of integration, so I'll say cosine of x, cosine of x squared. And then my variable of integration is x, so I'm really integrating of x squared, dx, but it'll give the same value, comma, from one until two, and now let the calculator munch on it a little bit, and I get approximately 2.557. So this is approximately 2.55, 2.557. Let me make sure that I added the three. Yeah, three plus that definite integral from one to two. 2.557. And I just rounded that. So there you go.