009A Sample Final 3, Problem 10 - Revision history

MathAdmin: Created page with "Let $y=\tan(x).$ (a) Find the differential $dy$
2017-04-10T16:32:58Z

Created page with "<span class="exam">Let <math style="vertical-align: -5px">y=\tan(x).</math> <span class="exam">(a) Find the differential <math style="vertical-align: -4px">dy</ma..."

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<span class="exam">Let  <math style="vertical-align: -5px">y=\tan(x).</math>

<span class="exam">(a) Find the differential  <math style="vertical-align: -4px">dy</math>  of  <math style="vertical-align: -5px">y=\tan (x)</math>  at  <math style="vertical-align: -15px">x=\frac{\pi}{4}.</math>

<span class="exam">(b) Use differentials to find an approximate value for  <math style="vertical-align: -5px">\tan(0.885).</math>  Hint:  <math style="vertical-align: -15px">\frac{\pi}{4}\approx 0.785.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|What is the differential  <math style="vertical-align: -4px">dy</math>  of  <math style="vertical-align: -4px">y=x^2</math>  at  <math style="vertical-align: -1px">x=1?</math>
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        Since  <math style="vertical-align: -4px">x=1,</math>  the differential is  <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
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'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First, we find the differential  <math style="vertical-align: -4px">dy.</math>
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|Since  <math style="vertical-align: -5px">y=\tan x,</math>  we have
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       <math>dy\,=\,\sec^2 x\,dx.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Now, we plug  <math style="vertical-align: -15px">x=\frac{\pi}{4}</math>  into the differential from Step 1.
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|So, we get
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       <math>dy\,=\,\bigg(\sec\bigg(\frac{\pi}{4}\bigg)\bigg)^2\,dx\,=\,2\,dx.</math>
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|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|First, we find  <math style="vertical-align: -1px">dx.</math>  We have
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|       <math style="vertical-align: -1px">dx=0.885-\frac{\pi}{4}\approx 0.885-0.785=0.1.</math>
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|Then, we plug this into the differential from part (a).
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|So, we have
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       <math>dy\,=\,2(0.1)\,=\,0.2.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Now, we add the value for  <math style="vertical-align: -4px">dy</math>  to  <math style="vertical-align: -16px">\tan\bigg(\frac{\pi}{4}\bigg)</math>  to get an
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|approximate value of  <math style="vertical-align: -5px">\tan(0.885).</math>
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|Hence, we have
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       <math>\tan(0.885)\,\approx \, \tan\bigg(\frac{\pi}{4}\bigg)+0.2\,=\,1.2.</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|    '''(a)'''     <math style="vertical-align: -5px">dy=2\,dx</math>
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|    '''(b)'''     <math style="vertical-align: -1px">1.2</math>
|}
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