course Mth 272 îÖ~µçÅQüŇ²““öלzz¾k|ޱ¢Ìassignment #005
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15:54:36 5.1.12 integrate 3 t^4 dt and check by differentiation
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RESPONSE --> Antiderivative 3t^4= 3*(t^5/5) = 3/5(t^5) Derivative 3/5t^5= 3/5*5t^4= 3t^4, the original function confidence assessment: 3
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15:54:46 An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c. The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. **
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RESPONSE --> ok self critique assessment: 3
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15:55:38 5.1.20 (was 5.1.18) integrate v^-.5 dv and check by differentiation
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RESPONSE --> -.5 +1= 0.5 deriv v^.0.5= 0.5v^-0.5 antiderivative= v^(0.5)/(0.5) simplified= 2v^(0.5) +c checking work deriv 2v^(0.5)= 2*0.5v^-0.5= v^-0.5, the original function confidence assessment: 3
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15:56:30 An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c. The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. **
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RESPONSE --> ok, was it alright that i put my exponent in decimal form or does it need to be fraction? self critique assessment: 2
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15:57:07 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> no comments as of now confidence assessment: 3
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