Chapter 12: Q 29. (page 916)

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

Short Answer

Expert verified

$\begin{array}{rcl} f_{1} (g_{1} (t), g_{2} (t)) & = & (\sin t)^{2} + (\cos t)^{2} \\ = & 1 \end{array}$

Step by step solution

Step 1. Given information

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

Step 2. Explanation

$\begin{array}{rcl} f_{1} (x, y) & = & x^{2} + y^{2} \\ g_{1} (t) & = & \sin t \\ g_{2} (t) & = & \cos t \end{array}$

Put $x = g_{1} (t); y = g_{2} (t)$ in $f_{1} (x, y)$

$\begin{array}{rcl} f_{1} (g_{1} (t), g_{2} (t)) & = & {(g_{1} (t))}^{2} + {(g_{1} (t))}^{2} \\ f_{1} (g_{1} (t), g_{2} (t)) & = & (\sin t)^{2} + (\cos t)^{2} \\ = & 1 \end{array}$

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In exercise,g1(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)=⟨1+t,t-1⟩,r2(t)=t,t2,t3Either simplify the specified composition or explain why thecomposition cannot be formed.f1g1t,g2t

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