Difference between revisions of "009A Sample Final 3, Problem 10"

Latest revision as of 16:49, 2 December 2017

Let $y=\tan(x).$

(a) Find the differential $dy$ of $y=\tan(x)$ at $x={\frac {\pi }{4}}.$

(b) Use differentials to find an approximate value for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan(0.885).} Hint: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pi}{4}\approx 0.785.}

Solution

Detailed Solution

Return to Sample Exam

@@ Line 5: / Line 5: @@
 <span class="exam">(b) Use differentials to find an approximate value for &nbsp;<math style="vertical-align: -5px">\tan(0.885).</math>&nbsp; Hint: &nbsp;<math style="vertical-align: -15px">\frac{\pi}{4}\approx 0.785.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 3, Problem 10 Solution|'''<u>Solution</u>''']]
-|-
-|What is the differential  &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; of &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; at &nbsp;<math style="vertical-align: -1px">x=1?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -4px">x=1,</math>&nbsp; the differential is &nbsp;<math style="vertical-align: -4px">dy=2xdx=2dx.</math>
-|}
-'''Solution:'''
+[[009A Sample Final 3, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we find the differential &nbsp;<math style="vertical-align: -4px">dy.</math>
-|-
-|Since &nbsp;<math style="vertical-align: -5px">y=\tan x,</math>&nbsp; we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,\sec^2 x\,dx.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we plug &nbsp;<math style="vertical-align: -15px">x=\frac{\pi}{4}</math>&nbsp; into the differential from Step 1.
-|-
-|So, we get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,\bigg(\sec\bigg(\frac{\pi}{4}\bigg)\bigg)^2\,dx\,=\,2\,dx.</math>
-|-
-|
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we find &nbsp;<math style="vertical-align: -1px">dx.</math>&nbsp;  We have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">dx=0.885-\frac{\pi}{4}\approx 0.885-0.785=0.1.</math>
-|-
-|Then, we plug this into the differential from part (a).
-|-
-|So, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,2(0.1)\,=\,0.2.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we add the value for &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; to &nbsp;<math style="vertical-align: -16px">\tan\bigg(\frac{\pi}{4}\bigg)</math>&nbsp; to get an
-|-
-|approximate value of &nbsp;<math style="vertical-align: -5px">\tan(0.885).</math>
-|-
-|Hence, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\tan(0.885)\,\approx \, \tan\bigg(\frac{\pi}{4}\bigg)+0.2\,=\,1.2.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -5px">dy=2\,dx</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -1px">1.2</math>
-|}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 3, Problem 10"

Latest revision as of 16:49, 2 December 2017

Navigation menu

Search