009A Sample Final 2, Problem 4 - Revision history

Kayla Murray at 17:13, 1 December 2017

2017-12-01T17:13:35Z

2017-03-18T20:29:41Z

2017-03-08T00:03:36Z

2017-03-07T23:50:50Z

2017-03-06T21:57:33Z

2017-03-06T21:46:43Z

2017-02-19T01:54:45Z

2017-02-18T02:27:38Z

@@ Line 3: / Line 3: @@
 ::<span class="exam"><math style="vertical-align: -4px">3x^2+xy+y^2=5</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 2, Problem 4 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 42: / Line 42: @@
 \displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\
 &&\\
-& = & \displaystyle{\frac{1}{-3}.}
+& = & \displaystyle{-\frac{1}{3}.}
 \end{array}</math>
 |-

@@ Line 17: / Line 17: @@
 !Step 1: &nbsp;
 |-
+|We use implicit differentiation to find the derivative of the given curve.
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>xy'+2yy'=5-6x-y</math>
 |-
+|and
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{5-6x-y}{x+2y}.</math>
 |}
@@ Line 29: / Line 37: @@
 !Step 2: &nbsp;
 |-
+|Therefore, the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>&nbsp; is
 |-
+|
@@ Line 38: / Line 56: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=-\frac{1}{3}(x-1)-2.</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 !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 17: / Line 17: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 40: / Line 38: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
-::<span class="exam"><math>3x^2+xy+y^2=5</math> at the point <math>(1,-2)</math>
+::<span class="exam"><math style="vertical-align: -4px">3x^2+xy+y^2=5</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"