AP Calculus Exam with the TI-89: Tips, Tricks, and Programs

<p>Like many of you, I am taking the AP Calculus AB exam soon and was wondering if anyone knows any helpful calculator tricks or programs that might be useful on the test.</p>

<p>Remember, there is a <em>pre-set</em> curve on the calc exam so it's okay to share, haha.</p>

<p>wouldn't it be easier and not to mention more fulfilling if you learn the math instead of relying on programs and technology?</p>

<p>The 89 Titanium needs no improvement. I think you already have a huge advantage. You should probably put in some power series programs. Oh, and that Euler's approximation stuff is pretty useful too.</p>

<p>Power Series and Euler's Approximation are not on the AB test.... only the BC</p>

<p>Just learn how to use your 89. You don't need many programs (if any). </p>

<p>However, I would recommend you download Calculus Tools from the TI website (it's free for your calculator). It has some useful stuff. Mainly I pull it out for implicit differentiation I really don't feel like doing by hand (usually a second derivative or third of a really nasty relation).</p>

<p>What can the 89 do that the 83 can't? Will i be at a disadvantage if i take the AP exam or the Math IIC with a TI-83+?</p>

<p>The 89 is capable of symbolic integration/differentiation, solving equations (not just equations in 0= form) and giving all solutions, and returning precise, fractional answers ( pi/4 instead of .785), along with a bunch of other things. </p>

<p>For the Math IIC, though the 89 can be useful, the 83+ is adequte. With practice, its possible to get a 800 regardless of what calculuator you have. For the Calculus AP exam (at least the BC one), the 89 makes everything a lot easier. Instead of having to take derivitives by hand, the calculator can do it for you. My school let everyone taking Calculus borrow a TI-89 for the month prior to the exam so that we could become familar with it. Whatever calculator you choose to use, make sure that you know it well. Going into the Math IIC with a brand new 89 that you have no idea how to use will probably end up hurting, not helping you.</p>

<p>TOOLS!!! ur calculator dependent... :( <em>sigh</em></p>

<p>Or maybe some people just like being able to know the work they did is correct?</p>

<p>Seriously, what's on there already is sufficient enough. Just know how to use it. Limits, integrals, derivs, you know the whole drill. Gotta love solver too.</p>

<p>ohnoes... maybe SOME but the vast majority do it so they don't have to think...</p>

<p>Hmm. The only argument that a sane person can make against the TI-89 as a bad thing is if the user does not know how to perform the operations without a calculator. This of course is the instructor's fault: if answers without work are accepted, it only encourages using a calculator without knowing the abilities. Don't bash the calculator, bash the teachers who don't make you learn to work without one.</p>

<p>the ap people claim the 89 won't help anywhere, yet i've seen it helps numerous times on practice mc exams... where's the justice?</p>

<p>I feel cheated! I'm taking both the Calc BC and Math IIC and my school can only provided us with the 83+. If I would have known I would have just bought the 89 at the beginning of the year. :-( LoL too late now..</p>

<p>Do you guys know the partial derivative trick for doing implicit differentiation insanely fast? </p>

<p>say you have an equation like tan(y)-sin(e^(y*x))=1 and you want to find dy/dx.</p>

<p>move everything to the side so you get an expression for the form F(x,y)=0. In this case: </p>

<p>tan(y)-sin(e^(y<em>x))-1=0 (</em>)</p>

<p>Now differentiate this equation in respect to x while pretending y is a constant (this is called partial differentiation...a topic normally taught in multivariable calculus).</p>

<p>you get -y<em>e^(x</em>y)<em>cos(e^(x</em>y)). (1)</p>

<p>Now do the same to (*) except with respect to y: that is, differentiate with respect to y while pretending x is a constant:</p>

<p>(secy)^2-x<em>e^(x</em>y)<em>cos(e^(x</em>y)). (2)</p>

<p>Now, the trick is this: multiply (1) by -1 and divide it by (2). You get:</p>

<p>dy/dx= [-y<em>e^(x</em>y)<em>cos(e^(x</em>y)<em>(cosy)^2]/[x</em>e^(x<em>y)</em>cos(e^(x<em>y)</em>(cosy)^2-1]</p>

<p>Tada! a seemingly impossible implicit differentiation question can be really easy using this trick. Here's a generalization</p>

<p>If you have F(x,y)=0 and you want dy/dx, </p>

<p>dy/dx=-F<em>x/F</em>y where F_t denotes the partial derivative of F with respect to t. </p>

<p>Skeptical? Try this one. Given x^2+y^2=1, find dy/dx.</p>

<p>F(x,y):=x^2+y^2-1. now, pretending y is a constant, we have F<em>x=2x
pretending x is a constant, we have F</em>y=2y. </p>

<p>now, dy/dx=-(2x)/(2y)=-x/y. If you do the whole implicit differentiation thing, you'll see you end up with the same answer.</p>

<p>Cool, no?</p>

<p>does that always work?!?!?</p>

<p>yes it does, some law or other, i can't remember the name, it comes up when you take multi-variable calculus (3rd semester normally)</p>

<p>forgive me for my ignorance, but doesn't differentiating tan give sec^2?</p>

<p>yeah, it does and it looks to me like that was done right....unless I'm missing something</p>

<p>hard to look at an equation typed out like that</p>

<p>yeah sorry about that. there might be some errors but it should be correct. </p>

<p>it has to do with the chain rule and "exact" derivatives</p>