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# 2015 AP Calculus BC 2b

## Video transcript

Part B four zero is less than T is less than one there is a point on the curve at which the line tangent to the curve has a slope of two the line tangent to the curve has a slope of two at what time is the object at that point so the slope of the tangent line is two that means that the rate of change of Y with respect to X is equal to two well they don't go directly give us the derivative of Y with respect to X but they do give us the derivative of X with respect to T that's the derivative of X with respect to T and they give us the derivative of Y with respect to T the X component of the velocity function is the rate of change of X with respect to time and the y component of the velocity function is the rate of change of Y with respect to time and using those two we can figure out the rate of change of Y with respect to X if you were to take the derivative of Y with respect to T and divide it by the derivative of X with respect to T derivative of X with respect to T well if you if you for the sake of of I guess conceptually understanding it if you view the differentials the way that you would view traditional numbers and fractions well the DTS would cancel out and you'd be left with dy divided by DX or a little bit more formally you could go to the chain rule and you'd say all right the derivative of Y with respect to T is equal to the derivative of Y with respect to x times the derivative of X with respect to T this comes straight out of the chain rule so this is the chain rule right over here that is the chain rule and then if you divide both sides by the derivative of X with respect to T you're going to get that original expression right over here well how is this useful well we know it the derivative of Y with respect to T is we know the derivative of X with respect to t is we know them as functions of T and then we can set them equal to two and then use our calculators to solve for T so let's do that the derivative of Y with respect to T that is e to the point 5t so we have e to the 0.5 T and then we divide it by the derivative of X with respect to T so that's going to be the X component of the velocity function so cosine of T squared and so this is the derivative of Y with respect to X and we need to figure out at what T does this equal to 2 or if we or if we want to simplify this and in our calculator we need to set this up so it's it's some expression you know some function of X is equal to 0 so let me let me rearrange this equation so I have a bunch of things equaling equaling 0 let's see I could just subtract 2 from both sides or actually what I could do is I could multiply both sides times cosine of T squared and so I'll have e to the 0.5 T is equal to 2 cosine of T squared and then I can subtract this from both sides and I will get e to the 0.5 t minus 2 cosine of T squared is equal to 0 and now I could use the solver on my calculator to figure this out so let's get the calculator out and let's go to math whoops let me make sure it's on so a math and then go all the way down to the solver so select that so my equation is zero is equal to so I'm going to say e to the 0.5 and the variable that going to solve for I'm going to use X instead of T the same thing I'll get the same answer e to the 0.5 X all right and then I'm going to have minus 2 times cosine of x squared is equal to 0 all right so there you have it 0 is equal to e to the 0.5 X minus 2 cosine of x squared click enter and then we put our initial guess and they tell us that 40 is between 0 and 1 so maybe a good guess would be right in between so let's put point 5 there and then we press alpha at least on this calculator and then you see that little blue solve there that will actually solve it let it munch on it for a little bit and I get T is equal to zero point eight four zero zero point eight four is zero so T so T is approximately zero point eight four zero and we are done
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