AP Calculus Problems with HP Prime

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AP Calculus Problems with HP Prime G. T. Springer Email: gt.springer@hp.

1 AP Calculus Problems with HP Prime G. T. Springer Hewlett-Packard Calculators and Educational Software In this hands-on workshop, participants will explore a number of problems in Calculus using the HP Prime graphing calculator. First, we will look at a problem involving implicit differentiation using the CAS and the Advanced Graphing app. Then we will examine the concept of a slope function using the Geometry app. As time permits, we will look at 1 or other problems. Part 1: Implicit Differentiation The point whose coordinates are (,0) is a point on the ellipse x + x y + 3y 4 = 0. What is the equation of the line tangent to the ellipse through that point? Press! to open the App Library and select the Advanced Graphing app. The app opens in its Symbolic view, where you can enter up to 10 equations or inequalities. In V1, enter the equation for our ellipse. Tap the menu keys at the bottom to enter =, X, and Y. Tap or press E when you are done. Press P to see the graph of the ellipse. Press + to zoom in and drag to center the ellipse, as shown in the figure to the right. Verify that the point (, 0) is actually on the curve. Tap, enter for X and 0 for Y. Tap. Press + to zoom in on this point until the curve straightens out and resembles a straight line segment. You can also drag the display with your finger to scroll the window. Finally you can zoom in or out using a pinch gesture. Place two fingers on the display at the same time and move them apart to zoom in. A vertical or horizontal pinch will zoom in one dimension only, while any diagonal pinch will zoom in both dimensions.

$Press c and choose the fraction template. The current tracer coordinates are stored in X and Y. So we can estimate the slope of the 0 Y curve at (, 0) using the expression.$ Our X estimate for the slope is close to -1. We will now use implicit differentiation to find the slope exactly. Press C to open CAS view.

Our X estimate for the slope is close to -1. We will now use implicit differentiation to find the slope exactly. Press C to open CAS view.

Press ^ to view the help page for this command.

2 We will now estimate the slope of the tangent line. Press R to move the cursor one pixel column to the right. Press H to open Home view. Press c and choose the fraction template. The current tracer coordinates are stored in X and Y. So we can estimate the slope of the 0 Y curve at (, 0) using the expression. Our X estimate for the slope is close to -1. We will now use implicit differentiation to find the slope exactly. Press C to open CAS view. Press b to open the Toolbox Menus, tap, then press i m (IM) to jump to commands that start with those two letters. Scroll down to implicit_diff. Press ^ to view the help page for this command. Notice the menu keys: : opens the entire help tree : opens a menu of examples to paste into the CAS : page by page navigation : view related commands : close the help page We can see that the command takes an expression, followed by the variables for differentiation. Tap to close the help page. Tap to paste the command into the CAS. The CAS uses lower-case variable names, so use x and y here.

3 Enter the expression for our ellipse, followed by both x and y, and press E to see the result. Tap to simplify the expression. Press c and select the third template in the first row (called the where() command. Tap on the first square, then tap on our last result and tap to copy it into the square. Tap on the second square and enter x=, y=0. Press E to see the result: the slope is -1. to return to Symbolic view and enter the equation of the tangent line. The line whose slope is -1 and contains (,0) is y-0= -1(x-) or y= -x+. Enter this equation in V and press P to see the graph. You can enter the equation in either point-slope form or slopeintercept form. Press - to zoom back out so you can see the entire ellipse.

Extension Find any values of a for which the parabola = a ( y + 5) We have three equations available to us: 1. Y= -x+ (or x= -y+).

Let s solve just as we would by hand. We can start with x = a( y + 5) and substitute x= -y+ to get an equation in a and y: y + = a( y + 5).

a a Press c to open the Template menu and select the substitution (or where() command), as shown to the right.

4 Extension Find any values of a for which the parabola = a ( y + 5) We have three equations available to us: 1. Y= -x+ (or x= -y+). x = a( y + 5) x = a y + 5, x, y) = 3. Implicit_diff( ( ) 1 Press C to return to the CAS. Enter these 4 equations, as shown in the figure to the right. Let s solve just as we would by hand. We can start with x = a( y + 5) and substitute x= -y+ to get an equation in a and y: y + = a( y + 5). Then we can solve the last 1 equation for y ( y = ) and substitute again to a get an equation in a alone: = a + 5. a a Press c to open the Template menu and select the substitution (or where() command), as shown to the right. Tap the first square in the template, then tap on our third equation and tap. Tap on the second square in the template and then copy the second equation into that square. When you are done, press E. The result is an equation in a and y, as shown to the right. x is also tangent to our line y= -x+.

Return to the substitution template and do the second substitution, as shown to the right.

Tap to simplify our equation in a. Multiply both sides of the equation by 4a simply by pressing x u A a.

5 Return to the substitution template and do the second substitution, as shown to the right. We could solve the fourth equation for y using the CAS, but it is simpler to do that mentally and enter the result manually. The substitution command does the substitution, but does not simplify the result (depending on your CAS Settings). Tap to simplify our equation in a. Multiply both sides of the equation by 4a simply by pressing x u A a. Simplify the result. This equation is quadratic in a. Subtract 8a from both sides, then subtract from both sides. The steps and final result are shown to the right. We can now use the Quadratic Formula to solve for a. The results are shown to the right as well.

We could have solved our equation in a directly. Press b, tap, select Solve and tap solve.

6 Return to Symbolic view and enter these equations in V3 and V4. Press P to see the graphs. Without graphing technology such as this, it would be far too difficult to verify our results graphically! We could have solved our equation in a directly. Press b, tap, select Solve and tap solve. Copy our equation in a as the first argument, followed by the variable a (separated by a comma). Press E to see the vector containing the solutions.

You can perform the entire process in one step if you like. Use the solve() command with the set of 3 equations enclosed in curly braces as a system, followed by the vector of variables.

7 You can perform the entire process in one step if you like. Use the solve() command with the set of 3 equations enclosed in curly braces as a system, followed by the vector of variables. The figure to the right shows the exact expression and the result. The result shows that: When a = -1/10, the parabola is tangent to the line at the point (-3, 5) When a= ½, the parabola is tangent to the line at the point (3, -1) These results can be verified in Plot view. The HP Prime CAS gives you extraordinary flexibility in exploring and solving problems exactly, while the Advanced Graphing app lets you verify your results graphically.

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HP Prime: The 10-Quicky Introduction Version 1.0. HP Prime: A Breakthrough in Mathematics Education Technology! HP Prime: A Breakthrough in Mathematics Education Technology! GT Springer, Solutions Architect HP Calculators and Educational Software gt.springer@hp.com This hands-on workshop is a basic introduction