- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Linearization
Problem 4.R.49
Textbook Question
Change in elevation The elevation h (in feet above the ground) of a stone dropped from a height of 1000 ft is modeled by the equation h(t) = 1000 - 16t², where t is measured in seconds and air resistance is neglected. Approximate the change in elevation over the interval 5 ≤ t ≤ 5.7 (recall that Δh ≈ h' (a) Δt).

1
First, identify the function that models the elevation of the stone: h(t) = 1000 - 16t^2.
To approximate the change in elevation, we need to find the derivative of h(t), which represents the rate of change of elevation with respect to time. Differentiate h(t) to get h'(t).
The derivative h'(t) = d/dt [1000 - 16t^2] = -32t. This represents the velocity of the stone at any time t.
Next, choose a point 'a' within the interval [5, 5.7] to evaluate the derivative. A common choice is the midpoint of the interval, a = (5 + 5.7) / 2 = 5.35.
Calculate the approximate change in elevation using the formula Δh ≈ h'(a) Δt, where Δt = 5.7 - 5. Substitute a = 5.35 and Δt = 0.7 into the formula to find the approximate change in elevation.
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