Finding Functions from Derivatives

In Exercis...

Question

Finding Functions from Derivatives

In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.

r'(t) = sec t tan t − 1, P(0, 0)

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the function that passes through the point P half of part 1, and has the derivative of 2 minus the cocC count of X multiplied by the cotangent of X using the second corollary of the mean value theorem. Now, how is this corollary going to help us to figure out the value of our function F of X? Well, recall that the coral basically tells us that if the derivative of F of X is equal to the derivative of some other function G of X at each point X in our open interval, OK. So in the interval AB. Then There exists a constant sea. Hm? Such that F of X will be equal to G of X plus C, OK? For all values of X in the interval, OK? So all X in the open interval AB. That is F minus G is a constant function on AB. So, for us to figure out FF X, first, let's try to figure out what our function G of X would be and then substitute the values in our point P into our function to solve for our constant. Then we'll be able to find our function F of X. Now remember earlier we were told that the derivative of F of X equals the derivative of G of X. So since the derivative of G of X is equal to 2 minus the coequant of X multiplied by the cotangent of X, what we need to do is to figure out what function after we differentiate it would give us this result. Well, if this is the derivative of G of X, then that means that G of X must be equal to 2 X plus the coc X because the derivative of 2 X is 2 and the derivative of the cocant of X is negative cocant X X. So in that case then. At the point P, OK. Half of I 1. Where we can tell that x is going to be equal to 1/2 of pi and F of X would be equal to 1. If we can substitute this into our formula for FFX equals GX + C, we should be able to solve for C and thus solve for FF X. Because no, in FFX equals G of X plus C if we substitute those values, then we'll get 1. To be equal to G of X, 2 X plus coc can of X where X is 1, so that's 2 multiplied by 1 plus the core C can of 1, oops sorry, remember X is half of pi, sorry. So it's 2 multiplied by 1/2 of pi plus the core C can of a half of pi, OK, plus C. Now what is this going to be? Well, We know that the cosicant of a half of pi, OK, would be equal to the reciprocal of the sign of a half of pi because the coequant is the reciprocal of the sine function. The sign of a half of pi is 1 and 1 divided by 1 equals 1. So in this case, our function is gonna be 1 equals 2 multiplied by 1/2 of pi, which is pi plus 1 + C. In this case, notice we have one on both sides of the equation, so we can cancel. Thus this implies that C equals negative pi. And if C equals negative pi, then that means our function F of X will be equal to G of X2 X plus the cocant of X minus pi, OK, because this is going to be C. Therefore, this is the function that passes through the point P half of 1 and has a derivative of 2 minus the coi can of X X. Thanks a lot for watching everyone. I hope this video helped.

Finding Functions from Derivatives

In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.

r'(t) = sec t tan t − 1, P(0, 0)

Key Concepts

Antiderivatives

Integration Techniques

Initial Conditions

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