Q 33. In exercise, letg1(t)=sint,聽g2(... [FREE SOLUTION]

Chapter 12: Q 33. (page 916)

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.
$f_{3} (g_{2} (t), g_{1} (t), g_{3} (t))$

Short Answer

Expert verified

Answer is $\frac{\cos t + \sin t}{\sin t + 1 - t}$

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

$\begin{array}{rcl} f_{3} (x, y, z) & = & \frac{x + y}{y + z} \\ g_{1} (t) & = & \sin t \\ g_{2} (t) & = & \cos t \\ g_{3} (t) & = & 1 - t \\ f_{3} (g_{2} (t), g_{1} (t), g_{3} (t)) & = & f_{3} (\cos t, \sin t, 1 - t) \\ = & \frac{\cos t + \sin t}{\sin t + 1 - t} \end{array}$

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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.f3g2t,g1t,g3t

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation

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