Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.2.31a

Chapter 3, Problem 3.2.31a

31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).
a. For the following functions s(t), find the instantaneous velocity function v(t). (Recall that the velocity function v is the derivative of the position function s.)
s(t)= −16t²+100t

Verified step by step guidance

Step 1: Identify the position function s(t) given in the problem, which is s(t) = -16t^2 + 100t.

Step 2: Recall that the instantaneous velocity function v(t) is the derivative of the position function s(t) with respect to time t.

Step 3: Differentiate the position function s(t) = -16t^2 + 100t with respect to t. Use the power rule for differentiation, which states that the derivative of t^n is n*t^(n-1).

Step 4: Apply the power rule to each term in s(t). The derivative of -16t^2 is -32t, and the derivative of 100t is 100.

Step 5: Combine the derivatives to find the velocity function v(t). Therefore, v(t) = -32t + 100.

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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 measures how the function's output changes as its input changes. In the context of motion, the derivative of the position function s(t) gives the instantaneous velocity v(t). This concept is fundamental in calculus as it provides a way to analyze rates of change.

Instantaneous Velocity

Instantaneous velocity refers to the velocity of an object at a specific moment in time. It is calculated as the derivative of the position function with respect to time. For the given position function s(t), finding v(t) involves applying differentiation to determine how fast the projectile is moving at any time t.

Quadratic Functions

A quadratic function is a polynomial function of degree two, typically expressed in the form s(t) = at² + bt + c. In this case, the position function s(t) = -16t² + 100t is a quadratic function, where the coefficients determine the shape of the parabola. Understanding the properties of quadratic functions is essential for analyzing projectile motion.

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Related Practice

Textbook Question

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

f(x) = 3x² - 4x; P(1, -1)

Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(s) = 4s³+3s; a= -3, -1

Textbook Question

79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>

a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)

x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

Textbook Question

A woman attached to a bungee cord jumps from a bridge that is 30 m above a river. Her height in meters above the river t seconds after the jump is y(t) = 15(1+e^-t cos t), for t ≥ 0.

Determine her velocity at t = 1 and t = 3.

Textbook Question

Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.

a. Show that the stones reach their high points at the same time.

Textbook Question

Consider the following cost functions.

a. Find the average cost and marginal cost functions.

C(x) = 1000+0.1x, 0≤x≤5000, a=2000