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'(θ) = 8 − csc²θ, P(π/4, 0)

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the function that passes through the point 3/4 of pi 0 and has the derivative of the square of the cant of X + 3, using the second corollary of the mean value theorem. Now how is this corollary going to help us? What does it say? Well, recall it basically tells us. That if the derivative of one function FX is equal to the derivative of another geoffX at each point, OK. In an open interval AB. Then there exists a constant sea. OK. Such that F of X is equal to G of X plus C, OK, for all values of X in the open interval. So, no, that really means that F minus G is a constant function on AB on our open interval. So, essentially, if we can figure out what the function G of X is, OK, then we can plug it into this expression that FX equals GX + C to help us solve for C and thus F of X. No, since we know that the derivative of FX equals geffi, then that means that the derivative of geafix will also be equal to the square of the c of X plus 3. So what would we differentiate to get this expression? Well, this must mean that G of X is going to be equal to the tangent of X plus 3 X because the derivative of the tangent of X is the square of the sant of X and the derivative of 3 X is 3. So now that we have this value of GFX, then at our point, OK, remember that, well, let me put this in red here. At our point P, where X was 3/4 of pi, and F of X is 0. OK, so, where X equals 3/4 of pi, and F of X equals 0. Then we can substitute it into our formula for FX equals G X + C to solve for constant C because C is constant, so it will be the same value regardless of whatever point. So no, this would imply then that F of X 0 is equal to G of X. The tangent of 3/4 of. Plus 3 multiplied by 3/4 of pi, K, plus C. Now we can go ahead and use this to solve for a sea. Now we know, OK, that the tangent of 3/4 of pie. Is going to be equal to the sine of 3/4 of pi divided by the cosine of 3/4 of pi because by definition the tangent is the ratio of the sign to the cosine. Now the sign of 3/4 of pi is 12 of root 2, and the cosine of 3/4 of pi is the negative value of a half of root 2. And when we divide those, then we get the tangent of 3/4 of pi to be -1. So this implies then that our equation will be 0 equals -1 + 3 multiplied by 3/4 of 9/4 of pi plus C. This implies that our constant C will be equal to 1 minus 9/4 of pi. I know that we have a value. Let me, let me write that a bit more clearly here. I know that we have a value for a C, OK, then this would imply. That if IX. Is going to be equal to g of X, the tangent of X plus 3 x plus or constant C which is 1 minus 9/4 of pi. Thus, this is the function that passes through the point P and has a derivative of the square of C X plus 3. 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'(θ) = 8 − csc²θ, P(π/4, 0)

Key Concepts

Antiderivatives

Trigonometric Integrals

Initial Conditions

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