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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.95b
Chapter 3, Problem 3.95b

Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2πr² + 2πrh.


b. How is dS/dt related to dh/dt if r is constant?

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Start by understanding the given formula for the total surface area of a right circular cylinder: \( S = 2\pi r^2 + 2\pi rh \). Here, \( r \) is the radius of the base, and \( h \) is the height of the cylinder.
Since \( r \) is constant, the term \( 2\pi r^2 \) is also constant, and its derivative with respect to time \( t \) is zero. Therefore, focus on differentiating the term \( 2\pi rh \) with respect to \( t \).
Apply the product rule to differentiate \( 2\pi rh \) with respect to \( t \). The product rule states that \( \frac{d}{dt}(uv) = u \frac{dv}{dt} + v \frac{du}{dt} \). Here, \( u = 2\pi r \) (a constant) and \( v = h \).
Differentiate \( 2\pi rh \) using the product rule: \( \frac{d}{dt}(2\pi rh) = 2\pi r \frac{dh}{dt} \). Since \( r \) is constant, \( \frac{dr}{dt} = 0 \), and the derivative simplifies to \( 2\pi r \frac{dh}{dt} \).
Thus, the rate of change of the surface area \( \frac{dS}{dt} \) with respect to time is directly proportional to the rate of change of the height \( \frac{dh}{dt} \), given by the expression \( \frac{dS}{dt} = 2\pi r \frac{dh}{dt} \).

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Key Concepts

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

Related Rates

Related rates are a concept in calculus that deals with the relationship between different rates of change. When two or more variables are related by an equation, the rate of change of one variable can be expressed in terms of the rate of change of another. In this context, we are interested in how the surface area of the cylinder changes with respect to time as the height changes, while keeping the radius constant.
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Intro To Related Rates

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of a function with respect to a variable. In this problem, we will differentiate the surface area equation with respect to time to find the relationship between the rates of change of surface area and height.
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Finding Differentials

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations that define one variable in terms of another without explicitly solving for one variable. In this case, we will apply implicit differentiation to the surface area equation to relate the change in surface area to the change in height, while treating the radius as a constant. This allows us to find the desired relationship between dS/dt and dh/dt.
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Finding The Implicit Derivative
Related Practice
Textbook Question

Consider the function


f(x) = { x² cos(2/x), x ≠ 0

0, x = 0


b. Determine f' for x ≠ 0.

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Textbook Question

Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365-day year. The equation that approximates the temperature on day x is


y = 37 sin[(2π/365)(x − 101)] + 25


and is graphed in the accompanying figure.


b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?


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Textbook Question

Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.


x√(1 + 2y) + y = x², P(1,0)

Textbook Question

Generalizing the Product Rule The Derivative Product Rule gives the formula


d/dx (uv) = u (dv/dx) + (du/dx) v


for the derivative of the product uv of two differentiable functions of x.


b. What is the formula for the derivative of the product u₁u₂u₃u₄ of four differentiable functions of x?

Textbook Question

Tolerance


b. About how accurately must the tank’s exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount?

Textbook Question

b. Show that


f(x) = { x² sin(1/x), x ≠ 0

0, x = 0

is differentiable at x = 0 and find f′(0).