   Chapter 14, Problem 22RE

Chapter
Section
Textbook Problem

Find all second partial derivatives of f.22. v = r cos(s + 2t)

To determine

To find: The second order partial derivative of the function v(r,s,t)=rcos(s+2t) .

Explanation

The given function is, v(r,s,t)=rcos(s+2t) .

Differentiate the given function with respect to r and obtain vr .

vr=r(rcos(s+2t))=cos(s+2t)r(r)=cos(s+2t)(1)

vr=cos(s+2t) (1)

Differentiate the equation (1) with respect to r and obtain the second order derivative, vrr .

2vr2=r(cos(s+2t))=0

Thus, vrr(r,s,t)=0 .

Differentiate the given function with respect to sand obtain vs .

vs=s(rcos(s+2t))=rs(cos(s+2t))=r[sin(s+2t)(1+0)]

vs=rsin(s+2t) . (2)

Differentiate the equation (2) with respect to s and obtain the second order derivative, vss .

2vs2=s(rsin(s+2t))=rs(sin(s+2t))=r[cos(s+2t)(1+0)]=rcos(s+2t)

Hence, vss(r,s,t)=rcos(s+2t) .

Differentiate the given function with respect to t and obtain vt .

vt=t(rcos(s+2t))=rt(cos(s+2t))=r[sin(s+2t)(0+2)]

vt=2rsin(s+2t) (3)

Differentiate the equation (3) with respect to t and obtain the second order derivative, vtt .

2vt2=t(2rsin(s+2t))=2rt(sin(s+2t))=2r[cos(s+2t)(0+2)]=4rcos(s+2t)

Hence, vtt(r,s,t)=4rcos(s+2t) .

Differentiate equation (1) with respect to s and obtain the partial derivative, vrs .

2vrs=s(cos(s+2t))=sin(s+2t)(1+0)=sin(s+2t)

Therefore, vrs(r,s,t)=sin(s+2t) .

Differentiate equation (2) with respect to r and obtain the partial derivative, vsr .

2vsr=r(rsin(s+2t))=sin(s+2t)r(r)=sin(s+2t)(1)=sin(s+2t)

Thus, vsr(r,s,t)=sin(s+2t)

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