<p>We have a test over 3-1 to 3-3 on Wednesday. I believe it covers everything in the chapter: critical points, intervals of increase/decrease, mean value theorem, Rolle’s theorem, absolute extrema, and something else (I forgot the name lol).</p>
<p>Well…My teacher broke chapter 3 into an A and B but we won’t have any quizes :(</p>
<p>Question: A point moves along the graph of y=x^2+4. As the point passes through (2,8), the x-coordinate increases at a rate of 2 units/sec. How fast is the distance between the point and the origin changing at this instant?</p>
<p>I got 8 units/sec as my answer, but I don’t feel confident about it at all. Could somebody walk me through this, or guide me to help online? I feel like a lost sheep.</p>
<p>Ugh… My teacher is really good, and I have a 100 in the class. His test reviews and notes are great to study… Except for tests. We had a test over chapter 2 today (differentiation - related rates) and he included a question on related rates of water evaporating in a cone that we had absolutely no understanding of. It wasn’t that we had no clue about related rates, but rather than we hadn’t been taught how to even make the inferences to point us in the right direction. He’s a very mathematical teacher… Not the best at teaching concepts. </p>
<p>Loving calculus so far though, really am. :)</p>
<p>I just wrote a great explanation @imwishing73 and I redirected and lost everything before I could post it -____-</p>
<p>@imwishing73 You solved for dy/dt, which is the rate at which y increases. The problem wants to know the rate at which the distance between the origin and point increases. You need to develop a relationship between x, y, and the distance between the point and origin. If you made a right triangle on the coordinate plane, you can see that to solve for the distance between point and origin, we use Pythagorean Theorem. </p>
<p>x^2 + y^2 = h^2 <------ With “h” being the distance between point and origin.
Now differentiate both sides with respect to time.</p>
<p>2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt) <------ We need to solve for dh/dt.</p>
<p>After rearranging and plugging in what you have (dx/dt = 2, and the point (2,8)), it can be seen that we are missing “h” and “dy/dt”. </p>
<p>By using the PT with x = 2 and y = 8, we see that h = sqrt(68) = 2sqrt(17)</p>
<p>Now we need to find dy/dt. YOU already did this with the original equation.
Differentiating the original equation with respect to time gives us:</p>
<p>dy/dt = 2x(dx/dt) <---- We known what both x and dx/dt are.</p>
<p>dy/dt = 2(2)(2) = 8 Also we just learned that:</p>
<p>h = 2sqrt(17)</p>
<p>Replug into the dh/dt equation that was made above:</p>
<p>[2(x)(dx/dt) + 2(y)(dy/dt)] / [2(h)] = dh/dt</p>
<p>[2(2)(2) + 2(8)(8)] / [2(2*(sqrt17))] = dh/dt</p>
<p>dh/dt = (136 / (4<em>(sqrt17))) = 2</em>(sqrt17)</p>
<p>dh/dt = <strong><em>8.246 units/sec</em></strong></p>
<p>When x is increasing at a rate of 2 units/sec, the rate at which the distance between the origin and (2,8) on the graph of y = x^2 + 4 is increasing at a rate of 8.246 units/sec.</p>
<p>Someone let me know if I did this wrong because we haven’t covered related rates in class yet. I just did an online lesson on it.</p>
<p>@biodontchem, THANK YOU SO MUCH. Your explanation was fantastic. Would you mind if I inboxed you with more questions later today?</p>
<p>Yes of course</p>
<p>More calc strugs 
If f(x)=(1+2x)^-5, then use the local linearization of function f at x=0 to approximate the value of f(0.3).</p>
<p>Using the formula f(x)=f(a)+f’(a)(x-a), I get -2. However, this isn’t one of the choices, and when I graph it in my calculator, the answer is closer to 0.9. I would assume that the multiple choice answer 0.8 is correct, but I want to know how to get there. Thanks!</p>
<p>My ap calc ab class is only on chapter 3 … which is derivates … chapter 4 is the application of derivates (related rates etc.) are we behind ? … we are finished chapter 3 and taking the test next week … so i guess im basically asking where is everyone else in there curriculm ?</p>
<p>Have a test tomorrow over concavity, inflection points, and 2nd derivative stuff.</p>
<p>Today in class, we discussed the chain rule… I feel like we’re kind of behind…</p>
<p>We just finished chapter 3. Our test is in 3 days. We never covered derivatives of inverse trig functions. Fail</p>
<p>Twiggy22- I think you might have the same book as me based on what you’re saying. Haha. Good luck!</p>
<p>That’s crazy. I think we’ll be starting integration after tomorrow’s test. Well, I didn’t really study, so yeah I’m probably not gonna do too well on tomorrow’s test haha. I’ll update you guys later.</p>
<p>Today we started Optimization/Applied Max and Min Problems. They are pretty similar to related rates problems, so they aren’t too bad. Where is everyone else at currently?</p>
<p>We just finished the first section of chapter 3 on extrema and critical points. Sounds like we’re behind!</p>
<p>Yeah, I probably failed the crap out of that test. Three people were crying in the next class after that test. Kinda crazy.</p>
<p>@elnamo … lol my calc book like its weird it has like a water ripple but the little water dots are different planets 
… yeah i think my class is behind … everybody is talking bout max and min integration inflection point … idk *** all tht is … so my class is behind … Lovely -__-</p>
<p>Does anyone know of any good calculus websites I can use to do practice problems?</p>