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

From Grad Wiki
Jump to navigation Jump to search
Line 14: Line 14:
 
!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>  
 
|}
 
|}
  
Line 33: Line 33:
 
!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>
 
 
|}
 
|}
  
Line 55: Line 55:
 
!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 17:25, 4 March 2017

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

between and about the -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 (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 x=y^3,~ \frac{dx}{dy}=3y^2.}
Now, we are going to integrate with respect to  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.}
Using the formula given in the Foundations section,
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{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  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}   is the surface area.
Step 2:  
Now, we use  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 u} -substitution.
Let  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 u=1+9y^4.}
Then,  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 du=36y^3dy}   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 \frac{du}{36}=y^3dy.}
Also, since this is a definite integral, we need to change the bounds of integration.
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 u_1=1+9(0)^4=1}   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 u_2=1+9(1)^4=10.}
Thus, we get
        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{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  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{dy}{dx}.}
Since  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}},}   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 \frac{dy}{dx}=\frac{27\sqrt{x}}{2}.}
Then, the arc 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 the curve 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 L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.}
Step 2:  
Then, 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 L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.}
Now, we use  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 u} -substitution.
Let  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 u=1+\frac{27^2x}{2^2}.}
Then,  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 du=\frac{27^2}{2^2}dx}   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 dx=\frac{2^2}{27^2}du.}
Also, since this is a definite integral, we need to change the bounds of integration.
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 u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}} 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 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}}}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009B_Sample_Final_2,_Problem_5&oldid=2985"