Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.8.10

Chapter 3, Problem 3.8.10

If r + s² + v³ = 12, dr/dt = 4, and ds/dt = –3, find dv/dt when r = 3 and s = 1.

Verified step by step guidance

Start by differentiating the given equation with respect to time t: \( \frac{d}{dt}(r + s^2 + v^3) = \frac{d}{dt}(12) \).

Apply the chain rule to differentiate each term: \( \frac{dr}{dt} + 2s \frac{ds}{dt} + 3v^2 \frac{dv}{dt} = 0 \).

Substitute the given values into the differentiated equation: \( 4 + 2(1)(-3) + 3v^2 \frac{dv}{dt} = 0 \).

Simplify the equation: \( 4 - 6 + 3v^2 \frac{dv}{dt} = 0 \), which simplifies to \( -2 + 3v^2 \frac{dv}{dt} = 0 \).

Solve for \( \frac{dv}{dt} \) by isolating it: \( 3v^2 \frac{dv}{dt} = 2 \), then \( \frac{dv}{dt} = \frac{2}{3v^2} \). Substitute the value of v when r = 3 and s = 1 to find \( \frac{dv}{dt} \).

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Key Concepts

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

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this problem, the equation r + s² + v³ = 12 involves multiple variables, and we need to differentiate with respect to time t to find the rate of change of v, denoted as dv/dt.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It is essential here because we are dealing with functions of multiple variables that change with respect to time. For instance, when differentiating s² with respect to t, we apply the chain rule: d(s²)/dt = 2s * (ds/dt).

Substitution of Known Values

After differentiating the equation, substituting known values is crucial to solve for the unknown rate of change. In this problem, we substitute r = 3, s = 1, dr/dt = 4, and ds/dt = -3 into the differentiated equation to find dv/dt. This step simplifies the equation, allowing us to isolate and solve for the desired rate.

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