   Chapter 12.2, Problem 41E

Chapter
Section
Textbook Problem

# Using Two MethodsIn Exercises 37 and 38, find (a) d d t [ r ( t ) ⋅ u ( t ) ] and (b) d d t [ r ( t ) × u ( t ) ] in two different ways.(i) Find the product first, then differentiate.(ii) Apply the properties of Theorem 12.2. r ( t ) = t i + 2 t 2 j + t 3 k ,     u ( t ) = t 4 k

(a)

To determine

To calculate: The expression ddt[r(t)u(t)] for the functions r(t)=ti+2t2j+t3k and u(t)=t4k.

Explanation

Given:

The functions r(t)=ti+2t2j+t3k and u(t)=t4k.

Formula used:

Let r and u be differentiable functions vector-valued functions of t, let w be a differentiable real valued functions of t, and let c be a scalar, then

1. ddt[cr(t)]=cr'(t)

2. ddt[r(t)×u(t)]=r(t)×u'(t)+r'(t)×u(t)

3. ddt[r(t)±u(t)]=r'(t)±u'(t)

4. ddt[r(t)u(t)]=r(t)u'(t)+r'(t)u(t)

5. ddt[r(t)u(t)]=r(t)u'(t)+r'(t)u(t)

If v1=x1i+y1j+z1k and v2=x2i+y2j+z2k are two vectors.

Then, dot product is given as:

v1v2=(x1i+y1j+z1k)(x2i+y2j+z2k)=x1x2+y1y2+z1z2

Calculation:

(i)

Consider the following functions:

r(t)=ti+2t2j+t3k and u(t)=t4k

Evaluate the dot product for r(t) and  u(t) and obtain the following result:

r(t)u(t)=(ti+2t2j+t3k)(t4k)=t(0)+2t2(0)+t7=t7

Differentiate the dot product obtained with respect to t as follows:

ddt(r(t)u(t))=ddt(t7)=7t6

Hence, the value of ddt[r(t)u(t)] is 7t6

(b)

To determine

To calculate: The expression ddt[r(t)×u(t)] for the functions r(t)=ti+2t2j+t3k and u(t)=t4k.

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