Domanda del libro di testo
Higher-order derivatives Find f′(x),f′′(x), and f′′′(x).
f(x) = 1/x
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.11.30
Parabolic motion An arrow is shot into the air and moves along the parabolic path y=x(50−x) (see figure). The horizontal component of velocity is always 30 ft/s. What is the vertical component of velocity when (a) x=10 and (b) x=40? <IMAGE>
Guida verificata passo dopo passo1
First, understand that the path of the arrow is given by the equation y = x(50 - x). This is a parabolic equation representing the vertical position y as a function of the horizontal position x.
To find the vertical component of velocity, we need to determine the derivative of y with respect to time t, which is dy/dt. However, y is given as a function of x, so we will use the chain rule: dy/dt = (dy/dx) * (dx/dt).
The horizontal component of velocity is given as dx/dt = 30 ft/s. This is constant throughout the motion.
Next, find dy/dx by differentiating y = x(50 - x) with respect to x. This gives dy/dx = 50 - 2x.
Now, substitute dy/dx and dx/dt into the chain rule expression: dy/dt = (50 - 2x) * 30. This expression gives the vertical component of velocity as a function of x. Evaluate this expression for (a) x = 10 and (b) x = 40 to find the vertical component of velocity at these points.
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Parabolic Motion
Parabolic motion refers to the trajectory of an object that is influenced by gravity, resulting in a curved path. In this case, the motion of the arrow follows a parabolic equation, which describes how the vertical position (y) changes with respect to the horizontal position (x). Understanding this concept is crucial for analyzing the motion of projectiles and determining their velocity components.
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Percorso guidato
06:29
Derivatives Applied To Velocity
Velocity Components
Velocity is a vector quantity that has both magnitude and direction. In projectile motion, the total velocity can be broken down into horizontal and vertical components. The horizontal component remains constant (30 ft/s in this case), while the vertical component changes due to gravitational acceleration. Calculating these components at specific points along the trajectory is essential for understanding the arrow's motion.
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Percorso guidato
06:29Derivatives Applied To Velocity
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. In the context of this problem, differentiating the parabolic equation y = x(50 - x) with respect to x allows us to determine the vertical component of velocity, which is the derivative of the height with respect to time. This mathematical tool is vital for analyzing motion and finding instantaneous rates.
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05:53Finding Differentials
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73–78. {Use of Tech} Normal lines A normal line at a point P on a curve passes through P and is perpendicular to the line tangent to the curve at P (see figure). Use the following equations and graphs to determine an equation of the normal line at the given point. Illustrate your work by graphing the curve with the normal line. <IMAGE>
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