   Chapter 13.3, Problem 23E

Chapter
Section
Textbook Problem

# Use Theorem 10 to find the curvature.23. r ( t ) = 6 t 2 i +2 t j +2 t 3 k

To determine

To find: The curvature of r(t)=6t2i+2tj+2t3k using Theorem 10.

Explanation

Given data:

The vector function r(t)=6t2i+2tj+2t3k .

Formula used:

Consider the three three-dimensional vectors such as a=a1i+a2j+a3k and b=b1i+b2j+b3k .

Cross product of vectors:

Write the expression for cross product of vectors a and b (a×b) .

a×b=|ijka1a2a3b1b2b3|

a×b=(a2b3b2a3)i(a1b3b1a3)j+(a1b2b1a2)k (1)

Write the expression for curvature of vector function r according to theorem 10.

k(t)=|r(t)×r(t)||r(t)|3 (2)

Here,

r(t) is first derivative of vector function r(t) , and

r(t) is second derivative of vector function r(t) .

Write the expression for magnitude of vector a=a1i+a2j+a3k (|a|) .

|a|=a12+a22+a32 (3)

Here,

a1 , a2 and a3 are the x, y, and z-coordinates of vector respectively.

Write the vector equation.

r(t)=6t2i+2tj+2t3k

Apply differentiation with respect to t on both sides of equation.

r(t)=ddt(6t2i+2tj+2t3k)=ddt(6t2)i+ddt(2t)j+ddt(2t3)k=(26t)i+2(1)j+2(3t2)k {ddx(x)=1ddx(xn)=nxn1}

r(t)=26ti+2j+6t2k (4)

Find the value of |r(t)| by using equation (3).

|r(t)|=(26t)2+(2)2+(6t2)2=24t2+4+36t4=4(9t4+6t2+1)=2((3t2)2+2(3t2)(1)+12)

|r(t)|=29t4+6t2+1=2(3t2+1)2{(a+b)2=a2+b2+2ab}=2(3t2+1)

Apply differentiation with respect to t on both sides of equation (4)

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