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)

Guida verificata passo dopo passo

Step 1: Understand the problem. We need to find a function r(t) whose derivative is given as r'(t) = sec(t) tan(t) - 1, and the function must pass through the point P(0, 0).

Step 2: Integrate the derivative r'(t) to find r(t). The integral of sec(t) tan(t) is sec(t), and the integral of -1 is -t. Therefore, r(t) = ∫(sec(t) tan(t) - 1) dt = ∫sec(t) tan(t) dt - ∫1 dt.

Step 3: Calculate the integrals separately. The integral of sec(t) tan(t) dt is sec(t), and the integral of -1 dt is -t. Thus, r(t) = sec(t) - t + C, where C is the constant of integration.

Step 4: Use the point P(0, 0) to find the constant C. Substitute t = 0 into r(t) = sec(t) - t + C, which gives r(0) = sec(0) - 0 + C = 1 + C. Since r(0) = 0, we have 1 + C = 0, so C = -1.

Step 5: Write the final function. With C = -1, the function is r(t) = sec(t) - t - 1. This function has the given derivative and passes through the point P(0, 0).

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Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse the process of differentiation. To find a function from its derivative, you need to determine its antiderivative. This involves integrating the given derivative function, which in this case is r'(t) = sec t tan t − 1, to find r(t).

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Antiderivatives

Integration Techniques

Integration techniques are methods used to find antiderivatives. For the derivative r'(t) = sec t tan t − 1, recognizing standard integrals is crucial. The integral of sec t tan t is sec t, and the integral of -1 is -t. Applying these techniques helps in constructing the original function r(t).

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Introduction to Indefinite Integrals

Initial Conditions

Initial conditions are used to determine the constant of integration when finding an antiderivative. Given the point P(0, 0), you substitute t = 0 and r(t) = 0 into the integrated function to solve for the constant. This ensures the function passes through the specified point, providing a complete solution.

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