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)

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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).

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).

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.

Finding Functions from DerivativesIn 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)