Q 35. In exercise, letg1(t)=sint,聽g2(... [FREE SOLUTION]

Chapter 12: Q 35. (page 917)

In exercise, let
$g_{1} (t) = \sin t, g_{2} (t) = \cos t, g_{3} (t) = 1 - t, f_{1} (x, y) = x^{2} + y^{2}, f_{2} (x, y) = \frac{x^{2}}{y^{2}}, f_{3} (x, y, z) = \frac{x + y}{y + z}, r_{1} (t) = 鉄� 1 + t, t - 1 鉄�, r_{2} (t) = <t, t^{2}, t^{3}>$
Either simplify the specified composition or explain why the composition cannot be formed.
$r_{2} (g_{3} (t))$

Short Answer

Expert verified

Answer is $<1 - t, {(1 - t)}^{2}, {(1 - t)}^{3}>$

Step by step solution

Step 1. Given information

$g_{1} (t) = \sin t, g_{2} (t) = \cos t, g_{3} (t) = 1 - t, f_{1} (x, y) = x^{2} + y^{2}, f_{2} (x, y) = \frac{x^{2}}{y^{2}}, f_{3} (x, y, z) = \frac{x + y}{y + z}, r_{1} (t) = 鉄� 1 + t, t - 1 鉄�, r_{2} (t) = <t, t^{2}, t^{3}>$

Step 2. Explanation

$\begin{array}{rcl} r_{2} (t) & = & <t, t^{2}, t^{3}> \\ g_{3} (t) & = & 1 - t \\ r_{2} (g_{3} (t)) & = & r_{2} (1 - t) \\ = & <1 - t, {(1 - t)}^{2}, {(1 - t)}^{3}> \end{array}$

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Short Answer

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In exercise, letg1(t)=sint,g2(t)=cost,g3(t)=1-t,f1(x,y)=x2+y2,f2(x,y)=x2y2,f3(x,y,z)=x+yy+z,r1(t)=鉄�1+t,t-1鉄�,r2(t)=t,t2,t3Either simplify the specified composition or explain why the composition cannot be formed.r2g3t

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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