   Chapter 13.3, Problem 26E

Chapter
Section
Textbook Problem

# Graph the curve with parametric equations x = cos t, y = sin t, z = sin 5t and find the curvature at the point (1, 0. 0).

To determine

To find: The graph and curvature of the curve with parametric equations x=cost,y=sint,z=sin5t at point (1,0,0) .

Explanation

Given data:

Parametric equation of curve are x=cost,y=sint,z=sin5t .

Formula used:

Write the expression for curvature of curve r(t) (K) .

K=|r(t)×r(t)||r(t)|3 (1)

Here,

r(t) is first derivative of r(t) ,

r(t) is second derivative of r(t) , and

r(t) is third derivative of r(t) ,

Cross product of vectors:

Write the expression for cross product of vectors u(t) and v(t) (u(t)×v(t)) .

u(t)×v(t)=|ijku1(t)u2(t)u3(t)v1(t)v2(t)v3(t)|=[(u2(t)v3(t)v2(t)u3(t))],[(u1(t)v3(t)v1(t)u3(t))],[(u1(t)v2(t)v1(t)u2(t))]

Write the expression for curve r(t) .

r(t)=x,y,z

Substitute cost for x, sint for y, and sin5t for z,

r(t)=cost,sint,sin5t

Equate the components of r(t) with components of point (1,0,0) .

cost=1t=cos1(1)t=0

sint=0t=sin1(0)t=0

sin5t=05t=sin1(0)5t=0t=0

Hence, t value is 0.

Consider the range of t is 0t2π .

The values of parametric equations for various values in the range 0t2π are listed in Table 1.

Table 1

 t x y z 0 1 0 0 π6 0.86603 0.5 0.5 π3 0.5 0.8660 –0.8660 π2 0 1 1 π –1 0 0 3π2 –0.5 0.8660 –0.8660 5π6 –0.866 0.5 0.5 2π 1 0 0

Sketch the graph of curve using the listed values in Table 1 as shown in Figure 1.

Find the value of r(t) .

r(t)=ddtcost,sint,sin5t=ddt(cost),ddt(sint),ddt(sin5t)=sint,cost,5cos5t{ddx(sinx)=cosx,ddx(cosx)=sinx,ddx(sinax)=acosax}

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

r(t)=ddtsint,cost,5cos5t=ddt(sint),ddt(cost),5ddt(cos5t)=cost,sint,25sin5t{ddx(sinx)=cosx,ddx(cosx)=sinx,ddx(cosax)=asinax}

Find the value of r(t)×r(t)

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