Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises - Page 815: 32

Answer

$$\eqalign{ & = 2{e^t}{t^2}\left( {16{t^5} - 9 + 2{t^6} - 3t} \right){\bf{i}} \cr & + 4{e^{ - t}}{t^2}\left( {16{t^5} - 9 - 2{t^6} + 3t} \right){\bf{j}} \cr & + 2{e^{2t}}{t^2}\left( { - 16{t^5} + 9 - 4{t^6} + 6t} \right){\bf{k}} \cr} $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{v}}\left( t \right) = {e^t}{\bf{i}} + 2{e^{ - t}}{\bf{j}} - {e^{2t}}{\bf{k}} \cr & {\text{Find }}{\bf{v}}'\left( t \right) = {e^t}{\bf{i}} - 2{e^{ - t}}{\bf{j}} - 2{e^{2t}}{\bf{k}} \cr & \cr & {\text{Calculate the derivative }}\frac{d}{{dt}}\left[ {\left( {4{t^8} - 6{t^3}} \right){\bf{v}}\left( t \right)} \right] \cr & = {\bf{v}}\left( t \right)\frac{d}{{dt}}\left[ {\left( {4{t^8} - 6{t^3}} \right)} \right] + \left( {4{t^8} - 6{t^3}} \right)\frac{d}{{dt}}\left[ {{\bf{v}}\left( t \right)} \right] \cr & = {\bf{v}}\left( t \right)\left( {32{t^7} - 18{t^2}} \right) + \left( {4{t^8} - 6{t^3}} \right){\bf{v}}'\left( t \right) \cr & {\text{Replacing }}{\bf{v}}\left( t \right){\text{ and }}{\bf{v}}'\left( t \right) \cr & = \left( {{e^t}{\bf{i}} + 2{e^{ - t}}{\bf{j}} - {e^{2t}}{\bf{k}}} \right)\left( {32{t^7} - 18{t^2}} \right) \cr & \,\,\,\,\, + \left( {4{t^8} - 6{t^3}} \right)\left( {{e^t}{\bf{i}} - 2{e^{ - t}}{\bf{j}} - 2{e^{2t}}{\bf{k}}} \right) \cr & {\text{Multiplying}} \cr & {\text{ = }}\left( {32{t^7}{e^t} - 18{t^2}{e^t}} \right){\bf{i}} + \left( {64{t^7}{e^{ - t}} - 36{t^2}{e^{ - t}}} \right){\bf{j}} \cr & \,\,\,\, - \left( {32{t^7}{e^{2t}} - 18{t^2}{e^{2t}}} \right){\bf{k}} + \left( {4{t^8}{e^t} - 6{t^3}{e^t}} \right){\bf{i}} \cr & + \left( { - 8{t^8}{e^{ - t}} + 12{t^3}{e^{ - t}}} \right){\bf{j}} + \left( { - 8{t^8}{e^{2t}} + 12{t^3}{e^{2t}}} \right){\bf{k}} \cr & \cr & {\text{ = }}32{t^7}{e^t}{\bf{i}} - 18{t^2}{e^t}{\bf{i}} + 64{t^7}{e^{ - t}}{\bf{j}} - 36{t^2}{e^{ - t}}{\bf{j}} \cr & \,\,\,\, - 32{t^7}{e^{2t}}{\bf{k}} + 18{t^2}{e^{2t}}{\bf{k}} + 4{t^8}{e^t}{\bf{i}} - 6{t^3}{e^t}{\bf{i}} \cr & - 8{t^8}{e^{ - t}}{\bf{j}} + 12{t^3}{e^{ - t}}{\bf{j}} - 8{t^8}{e^{2t}}{\bf{k}} + 12{t^3}{e^{2t}}{\bf{k}} \cr & \cr & {\text{Reducing like terms}} \cr & = \left( {32{t^7}{e^t} - 18{t^2}{e^t} + 4{t^8}{e^t} - 6{t^3}{e^t}} \right){\bf{i}} \cr & + \left( {64{t^7}{e^{ - t}} - 36{t^2}{e^{ - t}} - 8{t^8}{e^{ - t}} + 12{t^3}{e^{ - t}}} \right){\bf{j}} \cr & + \left( { - 32{t^7}{e^{2t}} + 18{t^2}{e^{2t}} - 8{t^8}{e^{2t}} + 12{t^3}{e^{2t}}} \right){\bf{k}} \cr & \cr & {\text{Factoring}} \cr & = 2{e^t}{t^2}\left( {16{t^5} - 9 + 2{t^6} - 3t} \right){\bf{i}} \cr & + 4{e^{ - t}}{t^2}\left( {16{t^5} - 9 - 2{t^6} + 3t} \right){\bf{j}} \cr & + 2{e^{2t}}{t^2}\left( { - 16{t^5} + 9 - 4{t^6} + 6t} \right){\bf{k}} \cr} $$
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