007A Sample Midterm 1 - Revision history

Kayla Murray: /* Problem 5 */

2017-11-02T19:50:13Z

Problem 5

Kayla Murray: /* Problem 5 */

2017-11-02T19:48:10Z

Problem 5

Kayla Murray: /* Problem 4 */

2017-11-02T19:44:01Z

Problem 4

Kayla Murray: /* Problem 3 */

2017-11-02T19:42:41Z

Problem 3

Kayla Murray at 19:38, 2 November 2017

2017-11-02T19:38:43Z

Kayla Murray: Created page with "text"

2017-11-02T19:37:09Z

Created page with "text"

New page

text

← Older revision		Revision as of 19:50, 2 November 2017
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	::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>		::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>

−	<span class="exam">where  <math style="vertical-align: ~~-4px~~">h</math>  is the height in centimeters and  <math style="vertical-align: 0px">m</math>  is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of  <math>m=85.</math>  What is the rate at  <math>h=170</math>  and  <math>h=190?</math>  Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?	+	<span class="exam">where  <math style="vertical-align: 0px">h</math>  is the height in centimeters and  <math style="vertical-align: 0px">m</math>  is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of  <math style="vertical-align: 0px">m=85.</math>  What is the rate at  <math style="vertical-align: -1px">h=170</math>  and  <math style="vertical-align: -1px">h=190?</math>  Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?

@@ Line 48: / Line 48: @@
 == [[007A_Sample Midterm 1,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==
-<span class="exam"> The displacement from equilibrium of an object in harmonic motion on the end of a spring is:
+<span class="exam"> To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula:
-::<span class="exam"><math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\sin(12t)</math>
+::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>
 '''Contributions to this page were made by [[Contributors|Kayla Murray]]'''

@@ Line 39: / Line 39: @@
 == [[007A_Sample Midterm 1,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==
-<span class="exam"> Find the derivatives of the following functions. Do not simplify.
+<span class="exam"> Find the derivatives of the following functions. '''Do not simplify.'''
 <span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>
@@ Line 45: / Line 45: @@
 <span class="exam">(b) &nbsp; <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>
-<span class="exam">(c) &nbsp; <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
+<span class="exam">(c) &nbsp; <math style="vertical-align: -8px">h(x)=\sqrt{x+\sqrt{x}}</math>
 == [[007A_Sample Midterm 1,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==

@@ Line 32: / Line 32: @@
 == [[007A_Sample Midterm 1,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==
-<span class="exam"> Let &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
+<span class="exam"> Let &nbsp;<math style="vertical-align: -5px">y=2x^2-3x+1.</math>
-<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
+<span class="exam">(a) Use the '''definition of the derivative''' to compute &nbsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math>
-<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
+<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -4px">y=2x^2-3x+1</math>&nbsp; at &nbsp;<math style="vertical-align: -4px">(2,3).</math>
 == [[007A_Sample Midterm 1,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==

← Older revision		Revision as of 19:38, 2 November 2017
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−	text	+	'''This is a sample, and is meant to represent the material usually covered in Math 7A for the midterm. An actual test may or may not be similar.'''
		+
		+	'''Click on the''' '''<span class="biglink" style="color:darkblue;"> boxed problem numbers </span> to go to a solution.'''
		+	<div class="noautonum">__TOC__</div>
		+
		+	== [[007A_Sample Midterm 1,_Problem_1\|<span class="biglink"><span style="font-size:80%"> Problem 1 </span></span>]] ==
		+	<span class="exam"> Find the following limits:
		+
		+	<span class="exam">(a) Find  <math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math>  provided that  <math style="vertical-align: -15px">\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]=5.</math>
		+
		+	<span class="exam">(b) Find  <math style="vertical-align: -14px">\lim _{x\rightarrow 0} \frac{\sin(4x)}{5x} </math>
		+
		+	<span class="exam">(c) Evaluate  <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>
		+
		+	== [[007A_Sample Midterm 1,_Problem_2\|<span class="biglink"><span style="font-size:80%"> Problem 2 </span>]] ==
		+	<span class="exam">Consider the following function  <math style="vertical-align: -5px"> f:</math>
		+	::<math>f(x) = \left\{
		+	\begin{array}{lr}
		+	x^2 & \text{if }x < 1\\
		+	\sqrt{x} & \text{if }x \geq 1
		+	\end{array}
		+	\right.
		+	</math>
		+
		+	<span class="exam">(a) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>
		+
		+	<span class="exam">(b) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>
		+
		+	<span class="exam">(c) Find  <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>
		+
		+	<span class="exam">(d) Is  <math style="vertical-align: -5px">f</math>  continuous at  <math style="vertical-align: -1px">x=1?</math>  Briefly explain.
		+
		+	== [[007A_Sample Midterm 1,_Problem_3\|<span class="biglink"><span style="font-size:80%"> Problem 3 </span>]] ==
		+	<span class="exam"> Let  <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
		+
		+	<span class="exam">(a) Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for  <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
		+
		+	<span class="exam">(b) Find the equation of the tangent line to  <math style="vertical-align: -5px">y=\sqrt{3x-5}</math>  at  <math style="vertical-align: -5px">(2,1).</math>
		+
		+	== [[007A_Sample Midterm 1,_Problem_4\|<span class="biglink"><span style="font-size:80%"> Problem 4 </span>]] ==
		+	<span class="exam"> Find the derivatives of the following functions. Do not simplify.
		+
		+	<span class="exam">(a)   <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>
		+
		+	<span class="exam">(b)   <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>
		+
		+	<span class="exam">(c)   <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
		+
		+	== [[007A_Sample Midterm 1,_Problem_5\|<span class="biglink"><span style="font-size:80%"> Problem 5 </span>]] ==
		+	<span class="exam"> The displacement from equilibrium of an object in harmonic motion on the end of a spring is:
		+
		+	::<span class="exam"><math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\sin(12t)</math>
		+
		+	<span class="exam">where  <math style="vertical-align: -4px">y</math>  is measured in feet and  <math style="vertical-align: 0px">t</math>  is the time in seconds.
		+
		+	<span class="exam">Determine the position and velocity of the object when  <math style="vertical-align: -14px">t=\frac{\pi}{8}.</math>
		+
		+
		+	'''Contributions to this page were made by [[Contributors\|Kayla Murray]]'''