022 Sample Final A, Problem 13 - Revision history

MathAdmin at 18:41, 6 June 2015

2015-06-06T18:41:02Z

MathAdmin at 04:41, 5 June 2015

2015-06-05T04:41:00Z

MathAdmin: Created page with "Use differentials to find $dy$ given $y = x^2 - 6x, ~ x = 4, ~dx = -0.5.$ {..."

2015-06-05T04:38:15Z

Created page with "<span class="exam">Use differentials to find <math style="vertical-align: -4px">dy</math> given <math style="vertical-align: -4px">y = x^2 - 6x, ~ x = 4, ~dx = -0.5.</math> {..."

New page

<span class="exam">Use differentials to find <math style="vertical-align: -4px">dy</math> given <math style="vertical-align: -4px">y = x^2 - 6x, ~ x = 4, ~dx = -0.5.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|When we use differentials, we are approximating a value for a function by using the slope of the derivative. The idea is that given a distance <math style="vertical-align: 0px">dx</math> from a point <math style="vertical-align: 0px">x</math>, we can use <math style="vertical-align: -5px">f'(x)</math>, the slope of the tangent line, to find the rise, <math style="vertical-align: -4px">dy</math>. Recalling that we can write
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::<math>f'(x)\,=\,\frac{dy}{dx},</math>
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|the relation is
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::<math>dy\,=\,f'(x)\cdot dx,</math>
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|where we use a given <math style="vertical-align: 0px">x</math>-value to evaluate <math style="vertical-align: -5px">f'(x)</math>.

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 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|By the power rule, we have
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::<math>f'(x) \,=\, 2x-6.</math>
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|We need to evaluate this at the given value <math style="vertical-align: -1px">x=4</math>, so
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::<math>f'(4) \,=\, 2(4)-6\,=\,2.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|We use the values given and the result from step 1 to find
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::<math>dy\,=\,f'(x)\cdot dx\,=\,2(-0.5)\,=\,-1.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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::<math>dy\,=\,-1.</math>
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[[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

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	<span class="exam">Use differentials to find <math style="vertical-align: -4px">dy</math> given <math style="vertical-align: -4px">y = x^2 - 6x, ~ x = 4, ~dx = -0.5.</math>		<span class="exam">Use differentials to find <math style="vertical-align: -4px">dy</math> given <math style="vertical-align: -4px">y = x^2 - 6x, ~ x = 4, ~dx = -0.5.</math>

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 <span class="exam">Use differentials to find <math style="vertical-align: -4px">dy</math> given <math style="vertical-align: -4px">y = x^2 - 6x, ~ x = 4, ~dx = -0.5.</math>
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 ::<math>dy\,=\,f'(x)\cdot dx,</math>
 |-
-|where we use a given <math style="vertical-align: 0px">x</math>-value to evaluate <math style="vertical-align: -5px">f'(x)</math>.
+|where we use the given specific <math style="vertical-align: 0px">x</math>-value to evaluate <math style="vertical-align: -5px">f'(x)</math>.

022 Sample Final A, Problem 13 - Revision history

MathAdmin at 18:41, 6 June 2015

MathAdmin at 04:41, 5 June 2015

MathAdmin: Created page with "Use differentials to find dy given y = x^2 - 6x, ~ x = 4, ~dx = -0.5. {..."

MathAdmin: Created page with "Use differentials to find $dy$ given $y = x^2 - 6x, ~ x = 4, ~dx = -0.5.$ {..."