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)

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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)

Bertha Stark · Accepted Answer

Step 1 To determine the parametric equations for a line passing through a given point, create the expression. &amp;nbsp;(x0,&amp;nbsp;y0,&amp;nbsp;z0)&amp;nbsp;are parallel to the vector&amp;nbsp;v=&amp;nbsp;&amp;lt;&amp;nbsp;a,&amp;nbsp;b,&amp;nbsp;c&amp;nbsp;&amp;gt;.x=x0&amp;nbsp;+&amp;nbsp;at,&amp;nbsp;y=y0&amp;nbsp;+&amp;nbsp;bt,&amp;nbsp;z=z0&amp;nbsp;+&amp;nbsp;ct&amp;nbsp;&amp;nbsp;Step 2 The parametric equations for the curves should be written as follows.x=e−8t&amp;nbsp;cos⁡(8t),&amp;nbsp;y=e−8t&amp;nbsp;sin⁡(8t),&amp;nbsp;z=e−8t&amp;nbsp;Step 3 Write the vector equation from the parametric equations of the curve as follows.&amp;nbsp;r(t)=⟨&amp;nbsp;e−8t&amp;nbsp;cos⁡(8t),&amp;nbsp;e−8t&amp;nbsp;sin⁡(8t),&amp;nbsp;e−8t⟩&amp;nbsp;Step 4 The derivative of the vector function r is the tangent vector of the curve (t). To find the derivative of the vector function, differentiate each component of the vector function.&amp;nbsp;r′(t)=⟨&amp;nbsp;ddt(e−8t&amp;nbsp;cos⁡(8t)),&amp;nbsp;ddt(e−8t&amp;nbsp;sin⁡(8t)),&amp;nbsp;ddt(e8t)⟩=&amp;nbsp;⟨&amp;nbsp;e−8t(−8&amp;nbsp;sin⁡(8t)&amp;nbsp;+&amp;nbsp;cos⁡(8t)(−8e−8t),&amp;nbsp;e−8t(8&amp;nbsp;cos⁡(8t))&amp;nbsp;+&amp;nbsp;sin⁡(8t)(−8e−8t,&amp;nbsp;−8e−8t)⟩=&amp;nbsp;⟨&amp;nbsp;−8e−8t&amp;nbsp;sin⁡(8t)&amp;nbsp;−&amp;nbsp;8e−8t&amp;nbsp;cos⁡(8t),&amp;nbsp;8e−8t&amp;nbsp;cos⁡&amp;nbsp;−&amp;nbsp;8e−8t&amp;nbsp;sin⁡(8t),&amp;nbsp;−8e−8t)⟩&amp;nbsp;Step 5 Given that the point&amp;nbsp;(1,&amp;nbsp;0,&amp;nbsp;1). That is,&amp;nbsp;1=e−8t&amp;nbsp;cos⁡(8t)&amp;nbsp;⇒&amp;nbsp;e−8t=1&amp;nbsp;or&amp;nbsp;cos⁡(8t)=1&amp;nbsp;⇒&amp;nbsp;t=00=e−8t&amp;nbsp;sin⁡(8t)&amp;nbsp;⇒&amp;nbsp;e−8t&amp;nbsp;≠&amp;nbsp;0&amp;nbsp;so&amp;nbsp;sin⁡(8t)=0&amp;nbsp;⇒&amp;nbsp;t=01=e−8t&amp;nbsp;⇒&amp;nbsp;t=0&amp;nbsp;As the specified point&amp;nbsp;(1,0,1)&amp;nbsp;corresponds to&amp;nbsp;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),&amp;nbsp;x=e−8(0)&amp;nbsp;cos⁡(8(0)),&amp;nbsp;y=e−8(0)&amp;nbsp;sin⁡(8(0)),&amp;nbsp;z=e−8(0)x=e0&amp;nbsp;cos⁡(0),&amp;nbsp;y=e0&amp;nbsp;sin⁡(0),&amp;nbsp;z=e−8(0)x=1,&amp;nbsp;y=0,&amp;nbsp;z=1&amp;nbsp;The point on the required line is&amp;nbsp;(1,&amp;nbsp;0,&amp;nbsp;1).&amp;nbsp;As the point on the required line is same as the specified point&amp;nbsp;(1,&amp;nbsp;0,&amp;nbsp;1),&amp;nbsp;the tangent vector is&amp;nbsp;r′(t)&amp;nbsp;at&amp;nbsp;t=0.&amp;nbsp;Step 6 Substitute 0 for t in &amp;lt;

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)

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