Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.68b

Chapter 6, Problem 6.1.68b

Variable gravity At Earth’s surface, the acceleration due to gravity is approximately g=9.8 m/s² (with local variations). However, the acceleration decreases with distance from the surface according to Newton’s law of gravitation. At a distance of y meters from Earth’s surface, the acceleration is given by a(y) = - g / (1+y/R)², where R=6.4×10⁶ m is the radius of Earth.

b. Use the Chain Rule to show that dv/dt = 1/2 d/dy(v²).

Verified step by step guidance

Recall that velocity $v$ is a function of time $t$, and position $y$ is also a function of time $t$. Therefore, $v = \frac{dy}{dt}$ and $v$ can be considered as $v(y(t))$.

Start with the expression $\frac{d}{dt}(v^2)$. Using the Chain Rule, this derivative can be written as $\frac{d}{dt}(v^2) = \frac{d}{dy}(v^2) \cdot \frac{dy}{dt}$.

Since $\frac{dy}{dt} = v$, substitute this into the expression to get $\frac{d}{dt}(v^2) = \frac{d}{dy}(v^2) \cdot v$.

Now, solve for $\frac{dv}{dt}$ by differentiating $v^2 = (v)^2$ with respect to $t$: $\frac{d}{dt}(v^2) = 2v \frac{dv}{dt}$.

Equate the two expressions for $\frac{d}{dt}(v^2)$: $2v \frac{dv}{dt} = \frac{d}{dy}(v^2) \cdot v$. Divide both sides by \$2v$ (assuming $v \neq 0$) to isolate $\frac{dv}{dt}$, yielding $\frac{dv}{dt} = \frac{1}{2} \frac{d}{dy}(v^2)$.

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Key Concepts

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

Chain Rule

The Chain Rule is a fundamental differentiation technique used to compute the derivative of a composite function. It states that if a variable depends on an intermediate variable, which in turn depends on another variable, the derivative is the product of the derivatives along the chain. In this problem, it helps relate derivatives with respect to time and position.

Relationship Between Velocity and Position

Velocity (v) is the rate of change of position (y) with respect to time (t), expressed as v = dy/dt. This relationship allows us to connect derivatives with respect to time and position, which is essential when applying the Chain Rule to rewrite dv/dt in terms of derivatives with respect to y.

Derivative of a Function Squared

When differentiating the square of a function, such as v², the power rule combined with the Chain Rule applies: d/dy(v²) = 2v dv/dy. Recognizing this helps transform expressions involving dv/dt into forms involving d/dy(v²), facilitating the proof requested in the question.

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