   Chapter 14, Problem 18RE

Chapter
Section
Textbook Problem

The speed of sound traveling through ocean water is a function of temperature, salinity, and pressure. It has been modeled by the functionC = 1449.2 + 4.6T − 0.055T2 + 0.00029T3 + (1.34 − 0.01T)(S − 35) + 0.016Dwhere C is the speed of sound (in meters per second), T is the temperature (in degrees Celsius), S is the salinity (the concentration of salts in parts per thousand, which means the number of grams of dissolved solids per 1000 g of water), and D is the depth below the ocean surface (in meters). Compute ∂C/∂T, ∂C/∂S, and ∂C/∂D when T = 10°C, S = 35 parts per thousand, and D = 100 m. Explain the physical significance of these partial derivatives.

To determine

To compute: The values of CT,CS and CD when T=10°C,S=35 parts per thousand, D=100m and explain the physical significance of these partial derivatives.

Explanation

Given:

The speed of sound traveling through ocean water is, C=1449.2+4.6T0.055T2+0.00029T3+(1.340.01T)(S35)+0.016D , where C is the speed of sound (in meters per second), S is the salinity (the number of grams of dissolved solids per1000g of water) and D is the depth below the ocean surface (in meters).

Calculation:

The given function is,

C=1449.2+4.6T0.055T2+0.00029T3+(1.340.01T)(S35)+0.016D .

Take the partial derivative with respect to T and obtain CT .

CT=T(1449.2+4.6T0.055T2+0.00029T3+(1.340.01T)(S35)+0.016D)={T(1449.2)+T(4.6T)T(0.055T2)+T(0.00029T3)+T[(1.340.01T)(S35)]+T(0.016D)}=0+4.6(1)0.055(2T)+0.00029(3T2)+(S35)(0.01)+0=4.60.11T+0.00087T20.01(S35)

Thus, CT=4.60.11T+0.00087T20.01(S35) (1)

Substitute T=10°C and S=35 in the equation (1) and obtain the value,

CT=4.60.11(10)+0.00087(10)20.01(3535)=4.61.1+0.0870=4.6871.1=3.587

Thus, the value of CT=3.587 . It represents that the speed of sound increases at a rate of 3.587 m/s per degree Celsius and when S and D are constants.

Take the partial derivative with respect to S and obtain CS .

CS=S(1449

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