   Chapter 12.2, Problem 71E

Chapter
Section
Textbook Problem

# Proof In Exercises 61-68, prove the property. In each case, assume r, u, and v are differentiable vector-valued functions of t in space, w is a differentiable real-valued function of t, and c is a scalar. d d t [ r ( w ( t ) ) ] = r ' ( w ( t ) ) w ' ( t )

To determine

To prove: The provided expression ddt[r(w(t))]=r(w(t)) w(t).

Explanation

Given:

Assume r, u, and v are differentiable vector-valued functions of t in space, w is a differentiable real-valued function of t and c is a scalar.

Formula used:

The product rule of differentiation is,

ddtf(g(t))=f'(g(t)) g'(t)

Proof:

Let the function r of t in space is,

r(t)=x(t)i+y(t)j+z(t)k

Where x(t), y(t) and z(t) are real valued differentiable function of t.

Let w is a differentiable real-valued function of t.

Then, the function r is,

r(w(t))=x(w(t))i+y(w(t))j+z(w(t))k

Now differentiate with respect to t both side of the above function.

ddt[r(w(t))]=ddt[x(w(t))i+y(w(t))j+z(w(t))k]=ddt[x(w(t))]i+ddt[y(w(t))]j+ddt[z

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