Typo fix (#524)

source/calculus/source/02-DF/01.ptx

@@ -112,7 +112,7 @@
     </activity>
 
  <definition xml:id = "def-inst-vel"> 
-     <p> The instanteous velocity at <m>t=a</m> is the limit as <m>h \to 0</m> of the difference quotient <m>\dfrac{f(a+h)-f(a)}{h}</m>. In the activity above the instantaneous velocity at <m>t=2</m> is given by the limit
+     <p> The instantaneous velocity at <m>t=a</m> is the limit as <m>h \to 0</m> of the difference quotient <m>\dfrac{f(a+h)-f(a)}{h}</m>. In the activity above the instantaneous velocity at <m>t=2</m> is given by the limit
      </p>
      <me>
      v(2) = \lim_{h\to 0} \dfrac{f(2+h)-f(2)}{h}</me>    
@@ -218,7 +218,7 @@
         <me>f'(a) = \lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}.</me>
     </definition>
 
-      <observation> <p>In <xref ref ="activity-ball1" />  and <xref ref ="activity-ball2" /> you studied a ball falling under gravity and estimated the instantaneous velocity as a limiting value of average velocities on smaller and smaller intervals. Drawing the corresponding secant lines, we see how the secant lines approximate better the tangent line, showing graphically what we previouly saw numericaly. Here is a Desmos animation showing the secant lines approaching the tangent line  <url href = "https://www.desmos.com/calculator/bzs1bxz7fa"/>.  </p></observation>
+      <observation> <p>In <xref ref ="activity-ball1" />  and <xref ref ="activity-ball2" /> you studied a ball falling under gravity and estimated the instantaneous velocity as a limiting value of average velocities on smaller and smaller intervals. Drawing the corresponding secant lines, we see how the secant lines approximate better the tangent line, showing graphically what we previously saw numerically. Here is a Desmos animation showing the secant lines approaching the tangent line  <url href = "https://www.desmos.com/calculator/bzs1bxz7fa"/>.  </p></observation>
 
 
     <activity xml:id = "meanings-derivative">
@@ -259,7 +259,7 @@
 
      <activity xml:id = "activity-differentiability-intuition">
               <introduction> 
-                <p> In this activity you will study the abolute value function <m>f(x)=|x|</m>. The absolute value function is a piecewise defined function which outputs <m>x</m> when <m>x</m> is positive (or zero) and outputs <m>-x</m> when <m>x</m> is negative. So the absolute value always outputs a number which is positive (or zero). Here is the graph of this function. </p>
+                <p> In this activity you will study the absolute value function <m>f(x)=|x|</m>. The absolute value function is a piecewise defined function which outputs <m>x</m> when <m>x</m> is positive (or zero) and outputs <m>-x</m> when <m>x</m> is negative. So the absolute value always outputs a number which is positive (or zero). Here is the graph of this function. </p>
                    <figure>
                     <image width="50%" xml:id="graph-absolute-value">
                       <sageplot>

source/calculus/source/02-DF/02.ptx

@@ -56,12 +56,12 @@
          <p>At any particular input <m>x=a</m>, the derivative function outputs <m>f'(a)</m>, the value of derivative at the point <m>x=a</m>. </p>
     </definition>
 
-    <remark xml:id="derivative-notation1">  <p>To specify the indendent variable of our function, we say that <m>f'(x)</m> is the derivative of <m>f(x)</m> with respect to <m>x</m>. For the derivative function of <m>y=f(x)</m> we also use the notation:</p>
+    <remark xml:id="derivative-notation1">  <p>To specify the independent variable of our function, we say that <m>f'(x)</m> is the derivative of <m>f(x)</m> with respect to <m>x</m>. For the derivative function of <m>y=f(x)</m> we also use the notation:</p>
         <me>f'(x)=y'(x)=\frac{dy}{dx}=\frac{df}{dx}.</me> 
         <p> The last type of notation is known as <em>differential</em> (or Leibniz) notation for the derivative.</p></remark>
 
    <remark xml:id="derivative-notation2"> 
-    <p>Notice that our notation for the derivative function is based on the name that we assign to the function along with our choice of notation for indendent and dependent variables.  For example, if we have a differentiable function <m>y=v(t)</m>, the derivative function of <m>v(t)</m> with respect to <m>t</m> can be written as <m>v'(t)=y'(t)=\dfrac{dy}{dt}=\dfrac{dv}{dt}</m>.</p>
+    <p>Notice that our notation for the derivative function is based on the name that we assign to the function along with our choice of notation for independent and dependent variables.  For example, if we have a differentiable function <m>y=v(t)</m>, the derivative function of <m>v(t)</m> with respect to <m>t</m> can be written as <m>v'(t)=y'(t)=\dfrac{dy}{dt}=\dfrac{dv}{dt}</m>.</p>
     </remark>
 
     <activity xml:id="activity-derivative-analytically-algebraic1">
@@ -213,7 +213,7 @@ The second derivative is the derivative of the derivative. It encodes informatio
        <activity xml:id = "DF2-activity-ball2">
               <introduction> 
                 <p>  In this activity you will study (again!) the velocity of a ball falling under gravity. A ball is tossed vertically in the air from a window. The height of the ball (in feet) is given by the formula
-<m> f(t) = 64-16(t-1)^2</m>, where <m>t</m> is the seconds after the ball is launched. Recall that in <xref ref="activity-ball1"/>, you used numerical methods to approxmiate the instantaneous velocity of <m>f(t)</m> to calculate <m>v(2)</m>! </p> </introduction>  
+<m> f(t) = 64-16(t-1)^2</m>, where <m>t</m> is the seconds after the ball is launched. Recall that in <xref ref="activity-ball1"/>, you used numerical methods to approximate the instantaneous velocity of <m>f(t)</m> to calculate <m>v(2)</m>! </p> </introduction>  
                   <task> <p> Using the limit definition of the derivative function, find the velocity function <m>v(t)=f'(t)</m>. </p></task>
              <task> <p> Using the velocity function <m>v(t)</m>, what is <m>v'(1)</m>, the instantaneous velocity at <m>t=1</m>? </p>   
             <ol marker="A." cols="2">

source/calculus/source/02-DF/04.ptx

@@ -336,7 +336,7 @@ In answering the following questions, be sure to explicitly denote which derivat
                 <li><m>r'''(x)=3 \, x \sin\left(x\right) - 9 \, \cos\left(x\right)</m></li>
                   <li><m>r^{(iv)}=3 \, x \cos\left(x\right) + 12 \, \sin\left(x\right)</m></li>
             </ul>
-            The functions involved repeat every four derivatives. So the twelth derivative should have the summand <m>3 \, x \cos\left(x\right)</m>. The second summand has a coefficient that increases by 3 every time, so the twelth derivative should have coefficient 36. Putting this together <m>r^{(xii)}=3 \, x \cos\left(x\right) + 36 \, \sin\left(x\right)</m>  </p> </li>
+            The functions involved repeat every four derivatives. So the twelfth derivative should have the summand <m>3 \, x \cos\left(x\right)</m>. The second summand has a coefficient that increases by 3 every time, so the twelfth derivative should have coefficient 36. Putting this together <m>r^{(xii)}=3 \, x \cos\left(x\right) + 36 \, \sin\left(x\right)</m>  </p> </li>
         </ol>
        <p/>
     </answer>-->

source/calculus/source/02-DF/05.ptx

@@ -120,7 +120,7 @@
                                  </activity>
 
   <activity xml:id="chain-rule-practice-graphs">
-   <introduction>  <p>Below you are given the graphs of two functions: <m>a(x)</m> and <m>b(x)</m>. Use the graphs to compute vaules of composite functions and of their derivatives, when possible (there are points where the derivative of these functions is not defined!). Notice that to compute the derivative at a point, you first want to find the derivative as a function of <m>x</m> and then plug in the input you want to study.  </p>
+   <introduction>  <p>Below you are given the graphs of two functions: <m>a(x)</m> and <m>b(x)</m>. Use the graphs to compute values of composite functions and of their derivatives, when possible (there are points where the derivative of these functions is not defined!). Notice that to compute the derivative at a point, you first want to find the derivative as a function of <m>x</m> and then plug in the input you want to study.  </p>
          <figure>
             <sidebyside widths="50% 50%">
                 <image xml:id="graph-chain-rule-practice-img1">
@@ -265,7 +265,7 @@
         <p> Find the derivative function <m>\dfrac{ds}{dt}</m> and check that <m>\dfrac{ds}{dt}|_{t=0} = 200</m>.  </p>
                </task>
        <task>
-        <p> According to this model, what is the maximum and minimum squirrel population? What is the fastest rate of increase and decrease of the squirrel population? When will these extremal scenarions occur?</p>
+        <p> According to this model, what is the maximum and minimum squirrel population? What is the fastest rate of increase and decrease of the squirrel population? When will these extremal scenarios occur?</p>
                </task>
                    </activity>
 

source/calculus/source/02-DF/07.ptx

@@ -12,7 +12,7 @@
   <subsection>
     <title>Activities</title>
     <observation xml:id="obs-df-implicit1">
-    Many of the equations that has been discussed so far fall under the category of an explicit equation.  An explicit equation is one in which the relationship between <m>x</m> and <m>y</m> is given explicitly, such as <m>y = f(x)</m>.  In this section we will examine when the relationship between <m>x</m> and <m>y</m> is given implicity. An implicit equation looks like <m>f(x,y) = g(x,y)</m> where both sides of the equation may depend on both <m>x</m> and <m>y</m>.
+    Many of the equations that has been discussed so far fall under the category of an explicit equation.  An explicit equation is one in which the relationship between <m>x</m> and <m>y</m> is given explicitly, such as <m>y = f(x)</m>.  In this section we will examine when the relationship between <m>x</m> and <m>y</m> is given implicitly. An implicit equation looks like <m>f(x,y) = g(x,y)</m> where both sides of the equation may depend on both <m>x</m> and <m>y</m>.
     </observation>
 
     <observation xml:id="obs-df-implicit2">

source/calculus/source/03-AD/01.ptx

@@ -49,7 +49,7 @@ Notice that this is the linear function with slope <m>f'(a)</m> and passing thro
     </activity>
 
     <definition xml:id="defn-motion"><p>
-        If a particle has position function <m>s = f(t)</m>, where <m>t</m> is measured in seconds and <m>s</m> is measured in meters, then the derivative of the position function tells us how the position is changing over time, so <m>f'(t)</m> gives us the (instantenous) velocity in meters per second. Also, the derivative of the velocity gives us the change in velocity over time, so so <m>f''(t)</m> gives us the (instantenous) acceleration in meters per second squared. Summarizing, 
+        If a particle has position function <m>s = f(t)</m>, where <m>t</m> is measured in seconds and <m>s</m> is measured in meters, then the derivative of the position function tells us how the position is changing over time, so <m>f'(t)</m> gives us the (instantaneous) velocity in meters per second. Also, the derivative of the velocity gives us the change in velocity over time, so so <m>f''(t)</m> gives us the (instantaneous) acceleration in meters per second squared. Summarizing, 
         <ul>
         <li><p><m>v(t) = f'(t)</m> is the velocity of the particle in <m>m/s</m>.</p></li>
         <li><p><m>a(t) = f''(t)</m> is the acceleration of the particle in <m>m/s^2</m>. </p></li>

source/calculus/source/03-AD/02.ptx

@@ -12,12 +12,6 @@
   <subsection>
     <title>Activities</title>
 
-<!--   <activity>
-    <introduction>
-      <p>Lorem ipsum dolor sit amet.</p>
-    </introduction>
-  </activity> -->
-
 
 <definition xml:id="defn-linear-approx"><p>
 The <em>linear approximation</em> (or tangent line approximation or linearization) of a function <m>f(x)</m> at <m>x=a</m> is the tangent line <m>L(x)</m> at <m>x=a</m>. In formulas, <m>L(x)</m> is the linear function

source/calculus/source/03-AD/04.ptx

@@ -69,7 +69,7 @@
 
 <observation>
     <p>
-    There can be some issues when trying to determine the global mimimum and maximum values of a function only using its graph. The Extreme Value Theorem will guarantee the existence of global extrema on a closed interval. Then we will see how to use derivatives to find algebraically the extrema of a function.
+    There can be some issues when trying to determine the global minimum and maximum values of a function only using its graph. The Extreme Value Theorem will guarantee the existence of global extrema on a closed interval. Then we will see how to use derivatives to find algebraically the extrema of a function.
     </p>
 </observation>
 
@@ -149,7 +149,7 @@
 
 <activity xml:id = "activity-ad-extrema-3">
     <statement>
-            <p>The Extreme Value Theorem (EVT) guarantees a global maximum and a global minumum for which of the following? </p>
+            <p>The Extreme Value Theorem (EVT) guarantees a global maximum and a global minimum for which of the following? </p>
             <ol marker="A." cols="2">
                 <li><m>f(x)=\dfrac{x^{2}}{x^{2}-4x-5}</m> on <m>[-5,0]</m>. </li>
                 <li><m>f(x)=\dfrac{x^{2}}{x^{2}-4x-5}</m> on <m>[0,4]</m>. </li>

source/calculus/source/03-AD/06.ptx

@@ -268,7 +268,7 @@
       </statement>
     </definition>
 <!-- 
- In <xref ref="AD5"/>, we used <xref ref="first-derivative-test"/> as a tool to determine when continuous functions had local maxima and local minima. In this section, we would like to develop a similar test that can identify where continous fucntions are concave up and concave down.  
+ In <xref ref="AD5"/>, we used <xref ref="first-derivative-test"/> as a tool to determine when continuous functions had local maxima and local minima. In this section, we would like to develop a similar test that can identify where continuous functions are concave up and concave down.  
      -->
     <activity xml:id="activity-derivative-concavity3">
         <introduction>

source/calculus/source/03-AD/07.ptx

@@ -109,7 +109,7 @@
             </p>
           </task>
            <task> <p>
-              A function <m>h(x)</m> thatis decreasing on <m>-3 \lt x \lt 3</m>,
+              A function <m>h(x)</m> that is decreasing on <m>-3 \lt x \lt 3</m>,
               concave up on <m>-3 \lt x \lt -1</m>,
               neither concave up nor concave down on <m>-1 \lt x \lt 1</m>,
               and concave down on <m>1 \lt x \lt 3</m>.

source/calculus/source/03-AD/09.ptx

@@ -67,7 +67,7 @@
     </task>
 
      </activity>
-    <activity><introduction><p> Compute the following limits using the limit defintion of the derivative at a point.</p> </introduction>
+    <activity><introduction><p> Compute the following limits using the limit definition of the derivative at a point.</p> </introduction>
     <task><p> <m>\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x}</m></p></task>
      <task><p> <m>\displaystyle \lim_{x\to 0} \frac{\tan(x)}{x}</m></p></task>
       <task><p> <m>\displaystyle \lim_{x\to 0} \frac{\cos(\frac{\pi}{3}+x) - \frac{1}{2} }{x}</m></p></task>  

source/calculus/source/04-IN/04.ptx

@@ -75,7 +75,7 @@
 
 <activity xml:id="activity-in-ivp-4">
     <statement>
-    <p>One of the applications of initial value problems is calculating the distance traveled from a point based on the velocity of the object.  Given that the velocity of the of an object in km/hr is approximated by <m>v(t) = \cos(t) + 1</m>, what is the approximate distance travelled by the object after 1 hour?</p>
+    <p>One of the applications of initial value problems is calculating the distance traveled from a point based on the velocity of the object.  Given that the velocity of the of an object in km/hr is approximated by <m>v(t) = \cos(t) + 1</m>, what is the approximate distance traveled by the object after 1 hour?</p>
         <ol marker="A." cols="2">
         <li><p>  <m>s(1) \approx 1</m> km </p></li>
         <li><p>  <m>s(1) \approx 0.1585</m> km </p></li> 

source/calculus/source/04-IN/05.ptx

@@ -14,7 +14,7 @@
 
 <activity xml:id="activity-in5-1">
     <introduction>
-        <p> Find the area beteween <m>f(x)=\frac{1}{2}x+2</m> and the <m>x</m>-axis from <m>x=2</m> to <m>x=6</m>. </p>
+        <p> Find the area between <m>f(x)=\frac{1}{2}x+2</m> and the <m>x</m>-axis from <m>x=2</m> to <m>x=6</m>. </p>
     </introduction>
 
     <figure xml:id="in5-figure-1">

source/calculus/source/04-IN/06.ptx

@@ -112,7 +112,7 @@
         <me>A(x) = \int_{g(x)}^{h(x)} f(t)\,dt</me>
             then 
             <me>A'(x) = \frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x),</me>
-            where <m>g(x)</m> and <m>h(x)</m> are continous differentiable functions.
+            where <m>g(x)</m> and <m>h(x)</m> are continuous differentiable functions.
         </p>
     </introduction>
 </theorem>

source/calculus/source/06-AI/05.ptx

@@ -98,7 +98,7 @@
         <activity xml:id="activity-AI4rectangularprism5pieces">
             <introduction>
                 <p>
-                    Suppose instead that we sliced the prism from <xref ref="activity-AI4rectangularprism"/> into 5 pices of height 4 meters.
+                    Suppose instead that we sliced the prism from <xref ref="activity-AI4rectangularprism"/> into 5 pieces of height 4 meters.
                     <figure>
                 <caption><m>10\times 10\times 20</m> prism sliced into 5 pieces.</caption>
                     <image xml:id="prism5pieces" width="50%">

source/calculus/source/07-CO/01.ptx

@@ -149,7 +149,7 @@ approximation <m>\sqrt 2\approx 0.707</m> to approximate coordinates as needed.)
             </p>
         </introduction>
         <task>
-            <p>Complete the folowing table.</p>
+            <p>Complete the following table.</p>
             <table>
                 <title>Chart of approximate <m>x</m> and <m>y</m> values</title>
                 <tabular>

source/calculus/source/07-CO/02.ptx

@@ -22,7 +22,7 @@ The coordinate on this graph at <m>t=2</m> is <m>(3,-3)</m>.
         <task>
             <p>
 Which of the following equations of <m>x,y</m> describes the graph of
-these paramteric equations?
+these parametric equations?
             </p>
             <ol marker="A." cols="2">
                 <li><p><m>y=2x(x+2)=2x^2+2x</m></p></li>

source/calculus/source/07-CO/03.ptx

@@ -83,7 +83,7 @@ Using the leg lengths and Pythagorean theorem, how long must the red dashed hypo
         <task>
             <p>
 Compared with the blue parametric curve connecting the same two points, is the red dashed line
-segement length an overestimate or underestimate?
+segment length an overestimate or underestimate?
             </p>
             <ol marker="A." cols="2">
                 <li><p>Overestimate: the blue curve is shorter than the red line.</p></li>
@@ -112,7 +112,7 @@ This formula will need to be modified to measure a curved path between two point
     <observation>
         <statement>
             <p>
-By approximating the curve by several (say <m>N</m>) segements connecting
+By approximating the curve by several (say <m>N</m>) segments connecting
 points along the curve, we obtain a better approximation than a single line segment.
 For example, the illustration shown in <xref ref="CO3-parametric-subdivide"/> gives
 three segments whose distances sum to about <m>7.6315</m>, while the actual length of the curve turns
@@ -148,7 +148,7 @@ in <xref ref="CO3-parametric-subdivide"/>?
         <task>
             <p>
 Let <m>\Delta L_1,\Delta L_2,\Delta L_3</m> describe the lengths of each
-of the three segements. Which expression describes the total length of these
+of the three segments. Which expression describes the total length of these
 segments?
             </p>
             <ol marker="A." cols="2">
@@ -178,7 +178,7 @@ Finally, we'll want to increase <m>N</m> from <m>3</m> so that it limits to <m>\
 What can we conclude when that happens?
             </p>
             <ol marker="A." cols="2">
-                <li><p>Each segment is infintely small.</p></li>
+                <li><p>Each segment is infinitely small.</p></li>
                 <li><p><m>\Delta x_i\to 0</m></p></li>
                 <li><p><m>\frac{\Delta x_i}{\Delta t_i}\to\frac{dx}{dt}</m></p></li>
                 <li><p>All of the above.</p></li>

source/calculus/source/07-CO/05.ptx

@@ -233,7 +233,7 @@ approximation <m>\sqrt 2\approx 0.707</m> to approximate coordinates as needed.)
             </p>
         </introduction>
         <task>
-            <p>Complete the folowing table.</p>
+            <p>Complete the following table.</p>
             <table>
                 <title>Chart of approximate <m>x</m> and <m>y</m> values</title>
                 <tabular>