   Chapter 13.3, Problem 58E

Chapter
Section
Textbook Problem

# The rectifying plane of a curve at a point is the plane that contains the vectors T and B at that point. Find the rectifying plane of the curve r(t) = sin t i + cos t j + tan t k at the point ( 2 / 2 , 2 / 2 , 1 ) .

To determine

To find: The rectifying plane of curve r(t)=sinti+costj+tantk at point (22,22,1) .

Explanation

Given data:

The curve is r(t)=sinti+costj+tantk and point is (22,22,1) .

Formula used:

Write the expression for tangent vector of a vector function r(t) (T(t)) .

T(t)=r(t)|r(t)| (1)

Here,

r(t) is first derivative of function r(t) .

Write the equation of plane with vector r(t)=r1,r2,r3 at point (x1,y1,z1) .

r1(xx1)+r2(yy1)+r3(zz1)=0 (2)

Consider the two three-dimensional vector functions such as u(t)=u1(t),u2(t),u3(t) and v(t)=v1(t),v2(t),v3(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 magnitude of vector a (|a|) .

|a|=a12+a22+a32

Here,

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

The rectifying plane contains vectors T and B , so the normal vector of rectifying plane is N since it is orthogonal to the vectors T and B .

Equate the components of r(t) with point (22,22,1) .

sint=22t=sin1(22)t=π4

cost=22t=cos1(22)t=π4

tant=1t=tan1(1)t=π4

Hence, the value of t is π4 .

Find the value of r(t) .

r(t)=ddt(sinti+costj+tantk)=ddt(sint)i+ddt(cost)j+ddt(tant)k=costi+(sint)j+sec2tk{ddx(sinx)=cosx,ddx(cosx)=sinx,ddx(tanx)=sec2x}=costisintj+sec2tk

Find the value of |r(t)| .

|r(t)|=(cost)2+(sint)2+(sec2t)2=cos2t+sin2t+sec4t=1+sec4t{cos2x+sin2x=1}

Substitute costisintj+sec2tk for r(t) and 1+sec4t for |r(t)| in equation (1),

T(t)=costisintj+sec2tk1+sec4t=cost1+sec4tisint1+sec4tj+sec2t1+sec4tk

Apply differentiation with respect to t on both sides of equation

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