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Let f(t)be a differentiable scalar function and r(t)be a differentiable vector function. Prove that ddt(f(t)r(t))=f'(t)r(t)+f(t)r'(t). (This is Theorem 11.11 (b).)

Short Answer

Expert verified

Ans:

ddt(f(t)r(t))=ddt(f(t)⟨x(t),y(t),z(t)⟩)=f'(t)r(t)+f(t)r'(t)

Step by step solution

01

Step 1. Given information: 

f(t)is differentiable scalar function and

r(t)is differentiable vector function.

02

Step 2. Proving :

Consider

ddt(f(t)r(t))=ddt(f(t)⟨x(t),y(t),z(t)⟩)

=ddt(⟨f(t)x(t),f(t)y(t),f(t)z(t)⟩)

=ddt(f(t)x(t))i+ddt(f(t)y(t))j+ddt(f(t)z(t))k

=f'(t)x(t)+f(t)x'(t)i+f'(t)y(t)+f(t)y'(t)j+f'(t)z(t)+f(t)z'(t)k

=f'(t)x(t)i+f'(t)y(t)j+f'(t)z(t)k+f(t)x'(t)i+f(t)y'(t)j+f(t)z'(t)k

=f'(t)(x(t)i+y(t)j+z(t)k)+f(t)x'(t)o+y'(t)j+z'(t)k

=f'(t)⟨x(t),y(t),z(t)⟩+f(t)x'(t),y'(t),z'(t)

=f'(t)r(t)+f(t)r'(t)

Thusddt(f(t)r(t))=f'(t)r(t)+f(t)r'(t)

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