009A Sample Final 1, Problem 7 - Revision history

MathAdmin at 16:13, 10 April 2017

2017-04-10T16:13:41Z

MathAdmin at 22:15, 18 April 2016

2016-04-18T22:15:36Z

MathAdmin at 19:17, 18 April 2016

2016-04-18T19:17:07Z

MathAdmin at 07:19, 5 March 2016

2016-03-05T07:19:17Z

MathAdmin: Created page with "A curve is defined implicitly by the equation :::::: $x^3+y^3=6xy.$ a) Using implicit differentiation, compute
2016-03-05T07:10:14Z

Created page with "A curve is defined implicitly by the equation ::::::<math>x^3+y^3=6xy.</math> a) Using implicit differentiation, compute <math..."

New page
A curve is defined implicitly by the equation

::::::<math>x^3+y^3=6xy.</math>

a) Using implicit differentiation, compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>.

b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
|-
|
::It would be  <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>  by the Product Rule.
|-
|'''2.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''3.''' What is the slope of the tangent line of a curve?
|-
|
::The slope is  <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Using implicit differentiation on the equation  <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
|-
|
::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we move all the  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  terms to one side of the equation.
|-
|So, we have
|-
|
::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
|-
|We solve to get  <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find the slope of the tangent line at the point  <math style="vertical-align: -5px">(3,3).</math>
|-
|We plug <math style="vertical-align: -5px">(3,3)</math>  into the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  we found in part '''(a)'''.
|-
|So, we get
|-
|
::<math>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  and a point.
|-
|Thus, we can write the equation of the line.
|-
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>  is
|-
|
::<math>y\,=\,-1(x-3)+3.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
|-
|'''(b)'''  <math>y=-1(x-3)+3</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 3: / Line 3: @@
 ::::::<math>x^3+y^3=6xy.</math>
-<span class="exam">a) Using implicit differentiation, compute &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
+::<span class="exam">a) Using implicit differentiation, compute &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
-<span class="exam">b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
+::<span class="exam">b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 49: / Line 49: @@
 ::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
 |-
-|We solve to get &nbsp;<math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
+|We solve to get
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is defined implicitly by the equation
-::::::<math>x^3+y^3=6xy.</math>
+::<math>x^3+y^3=6xy.</math>
-::<span class="exam">a) Using implicit differentiation, compute &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
+<span class="exam">(a) Using implicit differentiation, compute &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
-::<span class="exam">b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
+<span class="exam">(b) Find an equation of the tangent line to the curve &nbsp;<math style="vertical-align: -4px">x^3+y^3=6xy</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(3,3)</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
+|'''1.''' What is the result of implicit differentiation of &nbsp;<math style="vertical-align: -4px">xy?</math>
 |-
+|
-::It would be&thinsp; <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&thinsp; by the Product Rule.
+&nbsp; &nbsp; &nbsp; &nbsp; It would be &nbsp;<math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&nbsp; by the Product Rule.
 |-
 |'''2.''' What two pieces of information do you need to write the equation of a line?
 |-
+|
-::You need the slope of the line and a point on the line.
+&nbsp; &nbsp; &nbsp; &nbsp; You need the slope of the line and a point on the line.
 |-
 |'''3.''' What is the slope of the tangent line of a curve?
 |-
+|
-::The slope is&thinsp; <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; The slope is &nbsp;<math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
 |}
 '''Solution:'''
@@ Line 33: / Line 34: @@
 !Step 1: &nbsp;
 |-
-|Using implicit differentiation on the equation&thinsp; <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
+|Using implicit differentiation on the equation &nbsp;<math style="vertical-align: -4px">x^3+y^3=6xy,</math>&nbsp; we get
 |-
+|
-::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
 |}
@@ Line 42: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, we move all the &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; terms to one side of the equation.
+|Now, we move all the &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; terms to one side of the equation.
 |-
 |So, we have
 |-
+|
-::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
 |-
 |We solve to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
 |}
@@ Line 60: / Line 60: @@
 !Step 1: &nbsp;
 |-
-|First, we find the slope of the tangent line at the point &thinsp;<math style="vertical-align: -5px">(3,3).</math>
+|First, we find the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(3,3).</math>
 |-
-|We plug  <math style="vertical-align: -5px">(3,3)</math>&thinsp; into the formula for &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; we found in part '''(a)'''.
+|We plug  &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; into the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; we found in part (a).
 |-
 |So, we get
 |-
+|
-::<math>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 73: / Line 79: @@
 !Step 2: &nbsp;
 |-
-|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>&thinsp; and a point.
+|Now, we have the slope of the tangent line at &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; and a point.
 |-
 |Thus, we can write the equation of the line.
 |-
-|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>&thinsp; is
+|So, the equation of the tangent line at &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; is
 |-
+|
-::<math>y\,=\,-1(x-3)+3.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>y\,=\,-1(x-3)+3.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)'''&thinsp; <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
 |-
-|&nbsp;&nbsp; '''(b)'''&thinsp; <math>y=-1(x-3)+3</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>y=-1(x-3)+3</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]