# The derivative S ′ ( T ) and its unit. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 2.6, Problem 51E

(a)

To determine

## The derivative S′(T) and its unit.

Expert Solution

### Explanation of Solution

Given:

The quantity of oxygen that can dissolve in water depends on the temperature of the water.

Calculation:

The quantity of oxygen solubility S is depends on the water temperature T.

The quantity of oxygen solubility S(T) is a function of the water temperature T.

Note that, the derivative f(x) is the instantaneous rate of change of the function f(x) with respect to x

The derivative S(T) is the rate of change of quantity of oxygen solubility with respect to water temperature T.

The instantaneous rate of change is equal to S(T)=limΔx0ΔyΔt    here y=S(T).

Here Δy is measured in milligrams per liter and Δt is measured in centigrade.

Thus, the units are milligrams per liter per centigrade.

Therefore, the unit are (mg/L)/°C.

(b)

To determine

Expert Solution

### Explanation of Solution

Estimate the value of S(16).

Form given Figure, it is observed that the tangent line at 16 is passing through the point (8,12) and (24,8).

The slope of the tangent line to the curve at 16 as follows,

m=812248=416=0.25

Thus, the slope of the tangent line to the curve at 16 is, m=0.25(mg/L)/°C.

Note that, the slope of the tangent line to the curve at 16 is same S(16).

Therefore, S(16)=0.25(mg/L)/°C.

The derivative S(16)=0.25(mg/L)/°C means that the quantity of oxygen solubility decreasing at the rate of 0.25(mg/L)/°C as the temperature increasing post 16°C.

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