#2015 AP BC CALCULUS RESPONSE PLUS#

Wanted a coordinate, it would be 21, negative 3.226. But you could spilt it in to two integrals, it would just take some time.

An easier way to do this is, where you take the absolute value, which makes both areas positive, without you needing to split the area in to two integrals. Did I write that down right? I have a bad memory. You could use two integrals, where you would need to find the coordinate, where the two functions meet each other. That- so that's just this first part right over here. And this gives me- let theĬalculator think a little bit- 0.77356, and then to Integration is x, and I'm going to go from 0 to 3. To be sine of x squared- that's what I want toĮvaluate the definite integral of- from 0 to 3. Integral, and then I want to evaluate theĭefinite integral of this right over here. y of 3 is going to beĮqual to the integral from 0 to 3 of sine of x So it's going to be equal toĬ, which they tell us y of 0 is also equal to negative 4. You'll lose information from this constant. The derivative of this, you would get y of t, and To have to put a constant over here, because if you took Respect to x, and then one of the upper bound I didn't want to use a t hereĪnd then use a t, as well. And I'll just write itĪs sine of x squared dx. The fundamental theorem of calculus- y of t is equal This is equal to- and this is straight from Not even clear if youĬan do it analytically. Take the antiderivative of this right over here. So let's say y of t isĪntiderivative of this, this is not simple to

#2015 AP BC CALCULUS RESPONSE PLUS#

So if you want toįind x of 3, it's going to be 2 times 9 plus 3. So that tells usįind x of 3, you just evaluate it right over here. To 0 plus 0 plus c, which they're telling The whole thing is going to be equal to 0. Took the indefinite integral of this side, of 4t plus 1. T is going to be equal to the antiderivative So they give us x prime of t,Īnd they give us y prime of t. So we have sine of 9 dividedīy 13 gets as 0.0317. Get our calculator out and evaluate this. Just going to be- they tell us what dx/dt isĪs a function of time. So dy/dt at timeĮquals 3- well, this is dy/dt right over here. Time equals 3 and dx/dt at time equals 3. Slope at time equals 3, we just have to find dy/dt at

I wrote it this way is that they actually Sides, both the numerator and the denominator,īy dt over dt- which, if you do view them as T small changes in x, small changes in t. Little, small changes in y, small changes in, To deal with rigorously, but you can view them as very Is a little bit strange, especially when you want The rate of change of y with respect to x at that point. Tangent line is just going to be equal to Line tangent to the curve at this point is Y is equal to our slope is one fourth X plus and I could write it in any of these ways.Of the tangent line to the path of the particleĪt time t is equal to 3.

We could write it any of those ways, so the equation for the You add one fourth to both sides, you get B is equal to, we could either write itĪs one and one fourth, which is equal to five fourths, which is equal to 1.25. That's what it has inĬommon with the curve and so why is equal to one when X is equal to negative one, plus B and so we have one is equal to negative one fourth plus B. So includes this pointĪnd only that point. Now we need to solve for B and we know that point negative one comma one is on the line, so we can use that And so this is the same thing as three plus positive one, and so this is equal to one fourth and so the equation of our line is going to be Y is equal So three times one squared which is three, minus X, when Y is one, X is negative one, or when We'll see Y is, when X is negative one, Y is one, that sits on this curve. Going to be equal to DY/DX at that point, and we know that that's So the line's going to have a form Y is equal to MX plus B. Out the equation for the line if we know the slope of the line and we know a point that it goes through so that should be enough to figure out the equation of the line. Line tangent to the curve at the point negative one comma one. It can be shown that the derivative of Y with respect to X is equal Curve given by the equation Y to the third minus XY is equal to two.

2015 ap bc calculus response