Difference between revisions of "009B Sample Final 2, Problem 5"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|'''1.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is  
+
|'''1.''' The surface area &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; of a function &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; rotated about the &nbsp;<math style="vertical-align: -4px">y</math>-axis is given by
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>  
+
&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where &nbsp;<math style="vertical-align: -19px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
 
|-
 
|-
−
|'''2.''' The surface area &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; of a function &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; rotated about the &nbsp;<math style="vertical-align: -4px">y</math>-axis is given by
+
|'''2.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is  
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>  
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|We start by calculating &nbsp;<math>\frac{dx}{dy}.</math>  
+
|We start by calculating &nbsp;<math style="vertical-align: -16px">\frac{dx}{dy}.</math>  
 
|-
 
|-
−
|Since &nbsp;<math style="vertical-align: -13px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
+
|Since &nbsp;<math style="vertical-align: -17px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
 
|-
 
|-
−
|Now, we are going to integrate with respect to <math>y.</math>
+
|Now, we are going to integrate with respect to &nbsp;<math style="vertical-align: -3px">y.</math>
 
|-
 
|-
 
|Using the formula given in the Foundations section,
 
|Using the formula given in the Foundations section,
Line 47: Line 47:
 
&&\\
 
&&\\
 
& = & \displaystyle{2\pi \int_0^1 y^3 \sqrt{1+9y^4}~dy.}
 
& = & \displaystyle{2\pi \int_0^1 y^3 \sqrt{1+9y^4}~dy.}
 +
\end{array}</math>
 
|-
 
|-
−
|where &nbsp;<math>S</math>&nbsp; is the surface area.  
+
|where &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; is the surface area.  
−
\end{array}</math>
 
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we use <math>u</math>-substitution.  
+
|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math>u=1+9y^4.</math>
+
|Let &nbsp;<math style="vertical-align: -5px">u=1+9y^4.</math>
 
|-
 
|-
−
|Then, <math>du=36y^3dy</math> and <math>\frac{du}{36}=y^3dy.</math>
+
|Then, &nbsp;<math style="vertical-align: -5px">du=36y^3dy</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{36}=y^3dy.</math>
 
|-
 
|-
 
|Also, since this is a definite integral, we need to change the bounds of integration.
 
|Also, since this is a definite integral, we need to change the bounds of integration.
Line 65: Line 65:
 
|We have
 
|We have
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+9(0)^4=1</math> and <math>u_2=1+9(1)^4=10.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+9(0)^4=1</math>&nbsp; and &nbsp;<math>u_2=1+9(1)^4=10.</math>
 
|-
 
|-
 
|Thus, we get
 
|Thus, we get
Line 83: Line 83:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we calculate &nbsp;<math>\frac{dy}{dx}.</math>
+
|First, we calculate &nbsp;<math style="vertical-align: -15px">\frac{dy}{dx}.</math>
 
|-
 
|-
−
|Since <math>y=1+9x^{\frac{3}{2}},</math> we have
+
|Since &nbsp;<math style="vertical-align: -5px">y=1+9x^{\frac{3}{2}},</math>&nbsp; we have
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{27\sqrt{x}}{2}.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{27\sqrt{x}}{2}.</math>
 
|-
 
|-
−
|Then, the arc length &nbsp;<math>L</math>&nbsp; of the curve is given by  
+
|Then, the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of the curve is given by  
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.</math>
Line 101: Line 101:
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
 
|-
 
|-
−
|Now, we use &nbsp;<math>u</math>-substitution.  
+
|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let &nbsp;<math>u=1+\frac{27^2x}{2^2}.</math>
+
|Let &nbsp;<math style="vertical-align: -14px">u=1+\frac{27^2x}{2^2}.</math>
 
|-
 
|-
−
|Then, &nbsp; <math>du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math>dx=\frac{2^2}{27^2}du.</math>
+
|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">dx=\frac{2^2}{27^2}du.</math>
 
|-
 
|-
 
|Also, since this is a definite integral, we need to change the bounds of integration.
 
|Also, since this is a definite integral, we need to change the bounds of integration.
 
|-
 
|-
−
|We have &nbsp; <math>u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>
+
|We have  
 
|-
 
|-
−
|and &nbsp; <math>u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math> and &nbsp;<math>u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
 
|-
 
|-
 
|Hence, we now have
 
|Hence, we now have
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
−
|-
 
−
|
 
 
|}
 
|}
  

Revision as of 16:25, 4 March 2017

(a) Find the area of the surface obtained by rotating the arc of the curve

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y^3=x}

between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,0)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,1)} about the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} -axis.

(b) Find the length of the arc

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=1+9x^{\frac{3}{2}}}

between the points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,10)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (4,73).}

Foundations:  
1. The surface area  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S}   of a function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=f(x)}   rotated about the  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} -axis is given by

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S=\int 2\pi x\,ds,}   where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.}

2. The formula for the length  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L}   of a curve  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=f(x)}   where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\leq x \leq b}   is

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.}


Solution:

(a)

Step 1:  
We start by calculating  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dx}{dy}.}
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=y^{3},~{\frac {dx}{dy}}=3y^{2}.}
Now, we are going to integrate with respect to  
Using the formula given in the Foundations section,
we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {\int _{0}^{1}2\pi x{\sqrt {1+(3y^{2})^{2}}}~dy}\\&&\\&=&\displaystyle {2\pi \int _{0}^{1}y^{3}{\sqrt {1+9y^{4}}}~dy.}\end{array}}}
where    is the surface area.
Step 2:  
Now, we use  -substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=1+9y^{4}.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=36y^{3}dy}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{36}}=y^{3}dy.}
Also, since this is a definite integral, we need to change the bounds of integration.
We have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{1}=1+9(0)^{4}=1}   and  
Thus, we get
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {{\frac {2\pi }{36}}\int _{1}^{10}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}u^{\frac {3}{2}}{\bigg |}_{1}^{10}}\\&&\\&=&\displaystyle {{\frac {\pi }{27}}(10)^{\frac {3}{2}}-{\frac {\pi }{27}}.}\end{array}}}

(b)

Step 1:  
First, we calculate  
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=1+9x^{\frac {3}{2}},}   we have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {dy}{dx}}={\frac {27{\sqrt {x}}}{2}}.}
Then, the arc length    of the curve is given by
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=\int _{1}^{4}{\sqrt {1+{\bigg (}{\frac {27{\sqrt {x}}}{2}}{\bigg )}^{2}}}~dx.}
Step 2:  
Then, we have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=\int _{1}^{4}{\sqrt {1+{\frac {27^{2}x}{2^{2}}}}}~dx.}
Now, we use  -substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=1+{\frac {27^{2}x}{2^{2}}}.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {27^{2}}{2^{2}}}dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dx={\frac {2^{2}}{27^{2}}}du.}
Also, since this is a definite integral, we need to change the bounds of integration.
We have
        and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=1+{\frac {27^{2}(4)}{2^{2}}}=1+27^{2}.}
Hence, we now have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.}
Step 3:  
Therefore, we have
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{L} & = & \displaystyle{\frac{2^2}{27^2} \bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\ &&\\ & = & \displaystyle{\frac{2^3}{3^4} u^{\frac{3}{2}}\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\ &&\\ & = & \displaystyle{\frac{2^3}{3^4} (1+27^2)^{\frac{3}{2}}-\frac{2^3}{3^4} \bigg(1+\frac{27^2}{2^2}\bigg)^{\frac{3}{2}}.} \end{array}}


Final Answer:  
   (a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}}
   (b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2^3}{3^4} (1+27^2)^{\frac{3}{2}}-\frac{2^3}{3^4} \bigg(1+\frac{27^2}{2^2}\bigg)^{\frac{3}{2}}}

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