Find the parametric equations for the tangent line to the curve at the given location using the supplied parametric equations.x=1+2t,&nbsp;y=t3−t,&nbsp;z=t3+t;&nbsp;(3,0,2)

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Find the parametric equations for the tangent line to the curve at the given location using the supplied parametric equations.x=1+2t,&amp;nbsp;y=t3−t,&amp;nbsp;z=t3+t;&amp;nbsp;(3,0,2)

Heack1991 · Accepted Answer

We&#039;re given parametric equation&amp;nbsp;x=1+2t,&amp;nbsp;y=t3−t&amp;nbsp;and&amp;nbsp;z=t3+t&amp;nbsp;and we&#039;re asked to solve for the parametric equations of the tangent line to the curve at the point (3,0,2)We understand that the vector equation tr(t) is equal to given parametric equations.r(t)=&amp;lt;1+2t,t3−t,t3+t&amp;gt;Solve for&amp;nbsp;r′(t)&amp;nbsp;by differentiating each of the components of r(t) with respect to tr′(t)=&amp;lt;1t,3t2−1,3t2+1&amp;gt;The parameter value corresponding to&amp;nbsp;(3,0,2)&amp;nbsp;is t=1. Plug in t=1 into&amp;nbsp;r′(t)&amp;nbsp;to solve for&amp;nbsp;r′(1)r′(1)=&amp;lt;11,3(1)2−1,3(1)2+1&amp;gt;=&amp;lt;1,2,4&amp;gt;Consider the parametric equations for a line through the point&amp;nbsp;(x0,y0,z0)&amp;nbsp;and parallel to the direction vector&amp;nbsp;&amp;lt;a,b,c&amp;gt;&amp;nbsp;arex=x0+at&amp;nbsp;y=y0+bt&amp;nbsp;z=z0+ctSubstitute&amp;nbsp;(x0,y0,z0)=(3,0,2)&amp;nbsp;and&amp;nbsp;&amp;lt;a,b,c≥&amp;lt;1,2,4&amp;gt;&amp;nbsp;into x, y and z, respectively to solve for the parametric equations of the tangent line to the curvex=3+(1)t=3+t&amp;nbsp;y=(0)+(2)t=2t&amp;nbsp;z=(2)+(4)t=2+4t

user_27qwe · Accepted Answer

Step 1:Given parametric equations:x=1+2t,y=t3−t,z=t3+t.First, we find the derivatives of x, y, and z with respect to t:dxdt=ddt(1+2t)=2·12t=1t,dydt=ddt(t3−t)=3t2−1,dzdt=ddt(t3+t)=3t2+1.Next, we substitute t = 3 into these derivative equations to find the slope of the tangent line at the point (3, 0, 2):dxdt|t=3=13=33,dydt|t=3=3(3)2−1=26,dzdt|t=3=3(3)2+1=28.Therefore, the direction vector of the tangent line is (33,26,28).Step 2:Now, we can write the parametric equations for the tangent line in vector form:x=3+33t,y=26t,z=2+28t,where −∞&amp;lt;t&amp;lt;∞.Hence, the parametric equations for the tangent line to the curve at the point (3, 0, 2) are given by x=3+33t, y=26t, and z=2+28t, where t varies from −∞ to ∞.

star233 · Accepted Answer

Answer:x=3+ty=2tz=2+4tExplanation:First, let&#039;s find the value of t that corresponds to the point (3,0,2). We can set the x, y, and z coordinates equal to their respective expressions in terms of t and solve for t. So we have:1+2t=3 ⇒ 2t=2 ⇒ t=1 ⇒ t=1Now, we can calculate the derivatives of x, y, and z with respect to t:dxdt=ddt(1+2t)=2·12t=1tdydt=ddt(t3−t)=3t2−1dzdt=ddt(t3+t)=3t2+1Substituting t=1 into these derivatives, we get:dxdt|t=1=11=1dydt|t=1=3·12−1=2dzdt|t=1=3·12+1=4Now we have the direction vector of the tangent line, which is (1,2,4).Finally, we can write the parametric equations for the tangent line using the point-slope form. Let P(t) represent the parametric equations of the curve. The tangent line can be expressed as:x=3+1ty=0+2tz=2+4tTherefore, the parametric equations for the tangent line to the curve at the point (3,0,2) are:x=3+ty=2tz=2+4t

Find parametric equations for the tangent line to the curve

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