009A Sample Final 1, Problem 7 - Revision history

Kayla Murray at 22:55, 2 December 2017

2017-12-02T22:55:27Z

Kayla Murray at 20:21, 18 March 2017

2017-03-18T20:21:25Z

Kayla Murray at 16:29, 27 February 2017

2017-02-27T16:29:56Z

Kayla Murray at 00:53, 26 February 2017

2017-02-26T00:53:23Z

Kayla Murray at 01:48, 19 February 2017

2017-02-19T01:48:42Z

Kayla Murray at 01:44, 19 February 2017

2017-02-19T01:44:11Z

Grad at 20:32, 4 March 2016

2016-03-04T20:32:12Z

Grad: /* 4 */

2016-03-04T20:31:51Z

Grad: /* 3 */

2016-03-04T20:31:21Z

Grad: /* 2 */

2016-03-04T20:29:26Z

← Older revision		Revision as of 22:55, 2 December 2017
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	<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>.
		+	<hr>
		+	[[009A Sample Final 1, Problem 7 Solution\|'''<u>Solution</u>''']]

−	~~{\| 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>~~
−	\|}

		+	[[009A Sample Final 1, Problem 7 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''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>\begin{array}{rcl}~~
−	~~\displaystyle{m} & = & \displaystyle{\frac{3(3)^2-6(3)}{6(3)-3(3)^2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-\frac{9}{9}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-1.}~~
−	~~\end{array}</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\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Final_1\|'''<u>Return to Sample Exam</u>''']]

@@ Line 50: / Line 50: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<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 65: / Line 67: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<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 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 &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>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; 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?
@@ Line 23: / Line 23: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; 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>
 |}
@@ Line 34: / Line 34: @@
 !Step 1: &nbsp;
 |-
-|Using implicit differentiation on the equation&nbsp; <math style="vertical-align: -4px">x^3+y^3=6xy,</math> &nbsp; we get
+|Using implicit differentiation on the equation &nbsp;<math style="vertical-align: -4px">x^3+y^3=6xy,</math>&nbsp; we get
 |-
+|
@@ Line 60: / Line 60: @@
 |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 &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; 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
@@ Line 71: / Line 71: @@
 !Step 2: &nbsp;
 |-
-|Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math>&nbsp; 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>&nbsp; is
+|So, the equation of the tangent line at &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; is
 |-
+|

@@ Line 13: / Line 13: @@
 |-
+|
-::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&thinsp; <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&thinsp; 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&thinsp; <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
 |}
@@ Line 34: / 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 43: / 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;<math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
@@ Line 58: / Line 58: @@
 !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  <math style="vertical-align: -5px">(3,3)</math>&thinsp; 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>m\,=\,\frac{3(3)^2-6(3)}{6(3)-3(3)^2}\,=\,\frac{9}{-9}\,=\,-1.</math>
 |}
@@ Line 71: / Line 71: @@
 !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 <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 <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>
 |}
@@ Line 85: / Line 85: @@
 !Final Answer: &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>
 |-
-|'''(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>''']]

@@ Line 25: / Line 25: @@
 ::The slope is&thinsp; <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
 |}
 '''Solution:'''
@@ Line 79: / Line 80: @@
 ::<math>y\,=\,-1(x-3)+3.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ 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 &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;"