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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 40
Chapter 4, Problem 40

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)

Verified step by step guidance
1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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Related Practice
Textbook Question

Graphs and Graphing


Graph the curves in Exercises 33–42.


y = 𝓍³ (8―𝓍 )

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Textbook Question

Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = sin πt, s(0) = 0

342
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Textbook 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)

189
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Textbook Question

Finding Functions from Derivatives


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


g'(x) = 1 / x² + 2x, P(−1, 1)

257
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Textbook Question

Graphs and Graphing


Graph the curves in Exercises 33–42.

______

y = 𝓍√4 ― 𝓍²

175
views
Textbook Question

Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = 9.8t + 5, s(0) = 10

189
views