Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=e^{-8t} cos(8t), y=e^{-8t} sin(8t), z=e^{-8t}, (1, 0, 1)

Ernstfalld 2021-02-06 Answered
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=e8t cos(8t), y=e8t sin(8t), z=e8t, (1, 0, 1)
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Expert Answer

Bertha Stark
Answered 2021-02-07 Author has 96 answers

Step 1 Write the expression to find the parametric equations for a line through the point (x0, y0, z0) are parallel to the vector v= < a, b, c >.
x=x0 + at, y=y0 + bt, z=z0 + ct Step 2 Write the parametric equations of the curveas follows. x=e8t cos(8t), y=e8t sin(8t), z=e8t Step 3 Write the vector equation from the parametric equations of the curve as follows. r(t)= e8t cos(8t), e8t sin(8t), e8t Step 4 The tangent vector of the curve is the derivative of the vector function r(t). To find the derivative of the vector function, differentiate each component of the vector function. r(t)= ddt(e8t cos(8t)), ddt(e8t sin(8t)), ddt(e8t)
=  e8t(8 sin(8t) + cos(8t)(8e8t), e8t(8 cos(8t)) + sin(8t)(8e8t, 8e8t)
=  8e8t sin(8t)  8e8t cos(8t), 8e8t cos  8e8t sin(8t), 8e8t) Step 5 Given that the point (1, 0, 1). That is, 1=e8t cos(8t)  e8t=1 or cos(8t)=1  t=0
0=e8t sin(8t)  e8t  0 so sin(8t)=0  t=0
1=e8t  t=0 As the specified point (1,0,1) corresponds to t=0, consider the value of scalar parameter t as 0 and substitute in the parametric equations of the curve to obtain the point, which is on the required line. Substitute 0 for t in equation (2), x=e8(0) cos(8(0)), y=e8(0) sin(8(0)), z=e8(0)
x=e0 cos(0), y=e0 sin(0), z=e8(0)
x=1, y=0, z=1 The point on the required line is (1, 0, 1). As the point on the required line is same as the specified point (1, 0, 1), the tangent vector is r(t) at t=0. Step 6 Substitute 0 for t in

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