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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 56b
Chapter 3, Problem 56b

Assume that a particle’s position on the x-axis is given by


x = 3 cos t + 4 sin t,


where x is measured in feet and t is measured in seconds.


b. Find the particle’s velocity when t = 0, t = π/2, and t = π.

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To find the particle's velocity, we need to determine the derivative of the position function x(t) with respect to time t. The position function is given as x(t) = 3 cos(t) + 4 sin(t).
Differentiate x(t) with respect to t. The derivative of cos(t) is -sin(t), and the derivative of sin(t) is cos(t). Therefore, the derivative of x(t) is v(t) = -3 sin(t) + 4 cos(t).
Now, substitute t = 0 into the velocity function v(t) = -3 sin(t) + 4 cos(t) to find the velocity at t = 0.
Next, substitute t = π/2 into the velocity function v(t) = -3 sin(t) + 4 cos(t) to find the velocity at t = π/2.
Finally, substitute t = π into the velocity function v(t) = -3 sin(t) + 4 cos(t) to find the velocity at t = π.

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Key Concepts

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

Derivative

The derivative of a function represents the rate of change of the function with respect to a variable. In this context, the derivative of the position function x(t) with respect to time t gives the velocity of the particle. Calculating the derivative allows us to determine how the position changes over time, which is essential for finding the velocity at specific time points.
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Derivatives

Trigonometric Derivatives

Understanding the derivatives of trigonometric functions is crucial here, as the position function involves sine and cosine. The derivative of cos(t) is -sin(t), and the derivative of sin(t) is cos(t). Applying these rules to the position function x = 3 cos t + 4 sin t helps us find the velocity function, which is necessary to evaluate the velocity at given times.
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Derivatives of Other Inverse Trigonometric Functions

Evaluating Functions at Specific Points

Once the velocity function is derived, it must be evaluated at specific time points: t = 0, t = π/2, and t = π. This involves substituting these values into the velocity function to find the particle's velocity at these moments. This step is crucial for understanding the particle's motion at different times and requires careful substitution and simplification.
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Evaluating Composed Functions
Related Practice
Textbook Question

Quadratic approximations


[Technology Exercise] e. Find the quadratic approximation to h(x) = √(1 + x) at x = 0. Graph h and its quadratic approximation together. Comment on what you see.

Textbook Question

[Technology Exercise]


Graph y = 1/(2√x) in a window that has 0 ≤ x ≤ 2. Then, on the same screen, graph

y = (√(x + h) − √x)/h

for h = 1, 0.5, 0.1. Then try h = −1, −0.5, −0.1. Explain what is going on.

Textbook Question

Find the value of dy/dt at t = 0 if y = 3 sin 2x and x = t² + π.

Textbook Question

Assume that a particle’s position on the x-axis is given by


x = 3 cos t + 4 sin t,


where x is measured in feet and t is measured in seconds.


a. Find the particle’s position when t = 0, t = π/2, and t = π.

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Textbook Question

Derivative of y = |x| Graph the derivative of f(x) = |x|. Then graph y = (|x| − 0)/(x − 0) = |x|/x. What can you conclude?

Textbook Question

Quadratic approximations


d. Find the quadratic approximation to g(x) = 1/x at x = 1. Graph g and its quadratic approximation together. Comment on what you see