BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.8, Problem 33E
To determine

To find: The derivative of t with respect to c with the equation t=ln(3c+9c28c2) and write an interpretation statement based on the result.

Expert Solution

Answer to Problem 33E

The derivative of t with respect to c is given as [3[3c+9c28c]+(9c4)[3c(9c28c)+9c28c]].

Explanation of Solution

Given:

The given equation is as below.

t=ln(3c+9c28c2) (1)

Calculation:

Use chain rule to differentiate the equation (1).

Assume the variable u as (3c+9c28c2), and it implies that variable t as lnu.

Use the expression below.

dtdc=dtdu×dudcdtdc=1u×dudc

Substitute (3c+9c28c2) for u.

dtdc=13c+9c28c2×ddc(3c+9c28c2)

dtdc=2[3c+9c28c]×[32+12ddc(9c28c)] (1)

Refer equation (1), separate the term 9c28c.

Indicate it as f(c)=9c28c.

Differentiate f(c)=9c28c using the chain rule.

Assume the variable l as 9c28c, and it implies that variable f(c) as l.

Use the expression below.

df(c)dc=df(c)dl×dldcdf(c)dc=12l×dldc

Substitute 9c28c for l.

df(c)dc=12(9c28c)×ddc(9c28c)=12(9c28c)×(18c8)=(18c8)2(9c28c)=(9c4)(9c28c)

Substitute (9c4)(9c28c) for ddc(9c28c) in equation (1).

dtdc=2[3c+9c28c]×[32+12(9c4)(9c28c)]=[3[3c+9c28c]+1[3c+9c28c](9c4)(9c28c)]=[3[3c+9c28c]+(9c4)[3c(9c28c)+9c28c]]

Write the interpretation based on the result as below.

The rate of reaction will be fast with higher initial concentration value. The time taken for removing the unit concentration of the material will be lesser as the initial concentration increases.

The derivative of t with respect to c is given as [3[3c+9c28c]+(9c4)[3c(9c28c)+9c28c]].

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