Difference between revisions of "009B Sample Midterm 2, Problem 3"

Revision as of 10:21, 5 February 2017

A particle moves along a straight line with velocity given by:

v(t)=-32t+200

feet per second. Determine the total distance traveled by the particle

from time $t=0$ to time $t=10.$

Foundations:

Solution:

(a)

Step 1:

Step 2:

Final Answer:

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam"> Evaluate
+<span class="exam"> A particle moves along a straight line with velocity given by:
-::<span class="exam">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>
+::::::<math>v(t)=-32t+200</math>
-::<span class="exam">b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
+<span class="exam">feet per second. Determine the total distance traveled by the particle
+<span class="exam">from time <math>t=0</math> to time <math>t=10.</math>
@@ Line 9: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
+|
 |-
 |
-::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
 |-
 |
-::Thus, <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx=\int \sqrt{u}=\frac{2}{3}u^{3/2}+C=\frac{2}{3}(x^2+x)^{3/2}+C.</math>
 |}
@@ Line 24: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|We multiply the product inside the integral to get
+|
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>.
+|
 |}
@@ Line 32: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|We integrate to get
+|
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
+|
 |-
-|We now evaluate to get
+|
-|-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</math>.
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
-|-
-|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
 |-
-|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</math>.
+|
-|-
-|
 |}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We now have:
-|-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
-|-
-|So, we have
-|-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
-|}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
+|
-|-
-|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 2, Problem 3"

Revision as of 10:21, 5 February 2017

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