   Chapter 13, Problem 4RE

Chapter
Section
Textbook Problem

# Find parametric equations for the tangent line to the curve x = 2 sin t, y = 2 sin 2t, z = 2 sin 3t at the point (l, 3 , 2). Graph the curve and the tangent line on a common screen.

To determine

To find: The parametric equations for the tangent line to the curve with the parametric equations x=2sint,y=2sin2t,z=2sin3t at the point (1,3,2) and graph of a curve with the parametric equations x=2sint,y=2sin2t,z=2sin3t at the point (1,3,2) , and the tangent line on a common screen.

Explanation

Formula used:

Write the expression to find the parametric equations for a line through the point (x0,y0,z0) and parallel to the vector v=a,b,c .

x=x0+at,y=y0+bt,z=z0+ct (1)

Write the required differentiation formulae to find the tangent vector r(t) .

ddtsint=costddtsinnt=ncosnt

Write the parametric equations of the curve as follows:

x=2sint,y=2sin2t,z=2sin3t (2)

Write the vector equation from the parametric equations of the curve as follows:

r(t)=2sint,2sin2t,2sin3t

The tangent vector of the curve is the derivative of the vector function r(t) .

Calculation of the derivative of the vector function r(t) [r(t)] :

To find the derivative of the vector function, differentiate each component of the vector function.

Differentiate each component of the vector function r(t)=2sint,2sin2t,2sin3t as follows:

ddt[r(t)]=ddt(2sint),ddt(2sin2t),ddt(2sin3t)

Rewrite and compute the expression as follows:

r(t)=2ddt(sint),2ddt(sin2t),2ddt(sin3t)=2(cost),2(2cos2t),2(3cos3t)=2cost,4cos2t,6cos3t

As the scalar parameter t at π6 (t=π6) satisfies the vector function r(t)=2sint,2sin2t,2sin3t to attain the specified point (1,3,2) , consider the value of the scalar parameter t as π6 and substitute it in the parametric equations of the curve to obtain the point which is on the required line.

Calculation of the point on the required line:

Substitute π6 for t in equation (2),

x=2sin(π6),y=2sin[2(π6)],z=2sin[3(π6)]x=2(12),y=2(32),z=2(1)x=1,y=3,z=2

The point on the required line is (1,3,2)

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