009A Sample Final A, Problem 4 - Revision history

MathAdmin at 02:23, 28 March 2015

2015-03-28T02:23:25Z

2015-03-27T20:49:09Z

2015-03-27T05:57:36Z

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@@ Line 18: / Line 18: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>2\cdot y+2x\cdot y' = 2y +2x\cdot y'.</math><br>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 16: / Line 16: @@
 |which has as a derivative
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>2\cdot y+2x\cdot y' = 2y +2x\cdot y'.</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>2\cdot y+2x\cdot y' = 2y +2x\cdot y'.</math><br>
 |}
@@ Line 28: / Line 26: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>-3x^{2}-2y-2x\cdot y'+3y^{2}\cdot y'=0.</math>
 |-
-|From here, I would immediately plug in (1,1) to find ''y'' ':
+|From here, I would immediately plug in <math style="vertical-align: -22%">(1,1)</math> to find <math  style="vertical-align: -22%">y'</math>:
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -20%">-3-2-2y'+3y'=0</math>, or <math style="vertical-align: -20%">y' = 5.</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -20%">-3-2-2y'+3y'=0</math>, or <math style="vertical-align: -20%">y' = 5.</math><br>
 |}
@@ Line 44: / Line 40: @@
 |or in slope-intercept form
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>y=5x-4.</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>y=5x-4.</math><br>
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 |Since only two variables are present, we are going to differentiate everything with respect to <math style="vertical-align: 0%">x</math> in order to find an expression for the slope, <math style="vertical-align: -21%">m = y' = dy/dx</math>. Then we can use the point-slope equation form <math style="vertical-align: -21%;">y-y_{1} = m(x-x_{1})</math> at the point <math style="vertical-align: -21%">\left(x_1,y_1\right) = (1,1)</math> to find the equation of the tangent line.
 |-
-|Note that implicit differentiation will require the product rule <u>''and''</u> the chain rule.  In particular, differentiating <math style="vertical-align: -18%">2xy</math> must be treated as
+|Note that implicit differentiation will require the product rule <u>''and''</u> the chain rule.  In particular, differentiating <math style="vertical-align: -18%">2xy</math> can be treated as
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>(2x)\cdot (y),</math>

@@ Line 38: / Line 38: @@
 !Writing the Equation of the Tangent Line: &nbsp;
 |-
-|Now, we simply plug our values of <math style="vertical-align: -17%">x = y = 1</math> and <math style="vertical-align: 0%">m = 5</math> into the point-slope form to find the tangent line through <math style="vertical-align: -20%">(1,1)</math> is
+|Now, we simply plug our values of <math style="vertical-align: -17%">x = y = 1</math> and <math style="vertical-align: 0%">m = 5</math> &thinsp;into the point-slope form to find the tangent line through <math style="vertical-align: -20%">(1,1)</math> is
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>y-1=5(x-1),</math>

@@ Line 38: / Line 38: @@
 !Writing the Equation of the Tangent Line: &nbsp;
 |-
-|Now, we simply plug our values of <math style="vertical-align: -20%">x = y = 1</math> and <math style="vertical-align: 0%">m = 5</math> into the point-slope form to find the tangent line through <math style="vertical-align: -20%">(1,1)</math> is
+|Now, we simply plug our values of <math style="vertical-align: -17%">x = y = 1</math> and <math style="vertical-align: 0%">m = 5</math> into the point-slope form to find the tangent line through <math style="vertical-align: -20%">(1,1)</math> is
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>y-1=5(x-1),</math>

@@ Line 8: / Line 8: @@
 ! Foundations: &nbsp;
 |-
-|Since only two variables are present, we are going to differentiate everything with respect to ''x'' in order to find an expression for the slope, ''m'' = ''y'' ' = ''dy''/''dx''. Then we can use the point-slope equation form <math style="vertical-align: -21%;">y-y_{1} = m(x-x_{1})</math> at the point <math style="vertical-align: -21%">\left(x_1,y_1\right) = (1,1)</math> to find the equation of the tangent line.
+|Since only two variables are present, we are going to differentiate everything with respect to <math style="vertical-align: 0%">x</math> in order to find an expression for the slope, <math style="vertical-align: -21%">m = y' = dy/dx</math>. Then we can use the point-slope equation form <math style="vertical-align: -21%;">y-y_{1} = m(x-x_{1})</math> at the point <math style="vertical-align: -21%">\left(x_1,y_1\right) = (1,1)</math> to find the equation of the tangent line.
 |-
 |Note that implicit differentiation will require the product rule <u>''and''</u> the chain rule.  In particular, differentiating 2''xy'' must be treated as