009A Sample Final 3, Problem 5 - Revision history

Kayla Murray at 23:40, 2 December 2017

2017-12-02T23:40:31Z

Kayla Murray at 20:40, 18 March 2017

2017-03-18T20:40:24Z

Kayla Murray at 19:01, 6 March 2017

2017-03-06T19:01:05Z

Kayla Murray at 18:57, 6 March 2017

2017-03-06T18:57:23Z

Kayla Murray at 18:51, 6 March 2017

2017-03-06T18:51:46Z

Kayla Murray at 18:50, 6 March 2017

2017-03-06T18:50:40Z

Kayla Murray at 02:02, 19 February 2017

2017-02-19T02:02:03Z

Kayla Murray at 02:33, 18 February 2017

2017-02-18T02:33:34Z

Kayla Murray: Created page with "Compute ::a) $\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}$ ::b)
2017-02-18T02:15:51Z

Created page with "Compute ::a) <math style="vertical-align: -14px">\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}</math> ::b) <math st..."

New page
Compute

::a) <math style="vertical-align: -14px">\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}</math>

::b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin^2x}{3x}</math>

::c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math>

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'''Solution:'''

'''(a)'''

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'''(b)'''

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'''(c)'''

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[[009A_Sample_Final_3|'''Return to Sample Exam''']]

← Older revision		Revision as of 23:40, 2 December 2017
Line 1:		Line 1:
	<span class="exam"> Calculate the equation of the tangent line to the curve defined by  <math style="vertical-align: -4px">x^3+y^3=2xy</math>  at the point,  <math style="vertical-align: -5px">(1,1).</math>		<span class="exam"> Calculate the equation of the tangent line to the curve defined by  <math style="vertical-align: -4px">x^3+y^3=2xy</math>  at the point,  <math style="vertical-align: -5px">(1,1).</math>
		+	<hr>
		+	[[009A Sample Final 3, Problem 5 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|The equation of the tangent line to  <math style="vertical-align: -5px">f(x)</math>  at the point  <math style="vertical-align: -5px">(a,b)</math>  is~~
−	\|-
−	~~\|        <math style="vertical-align: -5px">y=m(x-a)+b</math>  where  <math style="vertical-align: -5px">m=f'(a).</math>~~
−	\|}

		+	[[009A Sample Final 3, Problem 5 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We use implicit differentiation to find the derivative of the given curve.~~
−	\|-
−	~~\|Using the product and chain rule, we get~~
−	\|-
−	~~\|        <math>3x^2+3y^2y'=2y+2xy'.</math>~~
−	\|-
−	~~\|We rearrange the terms and solve for  <math style="vertical-align: -5px">y'.</math>~~
−	\|-
−	~~\|Therefore,~~
−	\|-
−	~~\|        <math>3x^2-2y=2xy'-3y^2y'</math>~~
−	\|-
−	~~\|and~~
−	\|-
−	~~\|        <math>y'=\frac{3x^2-2y}{2x-3y^2}.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Therefore, the slope of the tangent line at the point  <math style="vertical-align: -5px">(1,1)</math>  is~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{m} & = & \displaystyle{\frac{3(1)^2-2(1)}{2(1)-3(1)^2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{3-2}{2-3}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-1.}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Hence, the equation of the tangent line to the curve at the point  <math style="vertical-align: -5px">(1,1)</math>  is~~
−	\|-
−	~~\|       <math>y=-1(x-1)+1.</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|        <math>y=-1(x-1)+1</math>~~
−	\|}
	[[009A_Sample_Final_3\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Final_3\|'''<u>Return to Sample Exam</u>''']]

@@ Line 47: / Line 47: @@
 |Hence, the equation of the tangent line to the curve at the point &nbsp;<math style="vertical-align: -5px">(1,1)</math>&nbsp; is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=-1(x-1)+1.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-1(x-1)+1.</math>
 |}
@@ Line 54: / Line 54: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(x)=-1(x-1)+1</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x-1)+1</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 35: / Line 35: @@
 !Step 2: &nbsp;
 |-
+|Therefore, the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(1,1)</math>&nbsp; is
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 44: / Line 54: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(x)=-1(x-1)+1</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 !Foundations: &nbsp;
 |-
+|The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
 |}

@@ Line 1: / Line 1: @@
-<span class="exam"> Calculate the equation of the tangent line to the curve defined by <math>x^3+y^3=2xy</math> at the point, <math>(1,1).</math>
+<span class="exam"> Calculate the equation of the tangent line to the curve defined by &nbsp;<math style="vertical-align: -4px">x^3+y^3=2xy</math>&nbsp; at the point, &nbsp;<math style="vertical-align: -5px">(1,1).</math>
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@@ Line 15: / Line 15: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 38: / Line 36: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 15: / Line 15: @@
 !Step 1: &nbsp;
 |-
+|We use implicit differentiation to find the derivative of the given curve.
 |-
+|Using the product and chain rule, we get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>3x^2+3y^2y'=2y+2xy'.</math>
 |-
+|We rearrange the terms and solve for &nbsp;<math style="vertical-align: -5px">y'.</math>
 |}

← Older revision		Revision as of 02:33, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">~~Compute~~	+	<span class="exam"> Calculate the equation of the tangent line to the curve defined by <math>x^3+y^3=2xy</math> at the point, <math>(1,1).</math>
−
−	~~::<span class="exam">a)~~ <math ~~style="vertical-align: -14px"~~>~~\lim_{x\rightarrow 4} \frac{\sqrt{~~x+~~5}-~~3~~}{x-4}</math>~~
−
−	~~::<span class="exam">b) <math style~~=~~"vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin^2x}{3x}~~</math>
−
−	~~::<span class="exam">c)~~ <math ~~style="vertical-align: -14px"~~>~~\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-~~1}</math>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"