<p>Someone posted my exact question on yahoo answers, but the answer some guy put is wrong (the answer is E, I just have no idea why).
Calculus</a> Question about Speed? - Yahoo! Answers</p>
<p>My other question is:</p>
<p>Which of the following is the solution to the differential equation dy/dx = 4x/y</p>
<p>I keep getting y= sqrt((4x^2)-12), but the answer has a negative before the sqrt.... help please?</p>
<p>Bumpppp please someone? D: I don’t think these are HARD questions, I just can’t seem to work them out</p>
<p>its all relative when you determine which graph is inc/dec f, f’, f’‘.
f is increasing when slope is pos, therefore f’ is positive.
f’ is increasing when f’’ is positive.</p>
<p>if u wanna find out if a function is inc/dec take its derivative.
In this case, speed is the absolute value of velocity, so take the derivative of velocity.</p>
<p>wait, so you have the ab calc 2003 released exam? :)</p>
<p>Yeah I have it in this book that I took from the teacher 
</p>
<p>But for my first question, how exactly do I find the derivative of the absolute value of the velocity?</p>
<p>nah u don’t need that, just find the derivative of velocity and u got the answer. no need to worry about absolute value for these types of problems.</p>
<p>For the dy/dx question, do this:</p>
<p>You have dy/dx = 4x/y. Separate variables so that y and dy is on one side, and x and dx are on the other. Then integrate.</p>
<p>y dy=4x dx</p>
<p>You should be able to do this; just solve for y or x or whatever you want to after you integrate.</p>
<p>The given point was (2,-2), but I keep getting the same answer I got before T_T</p>
<p>Yeah none of these replies have helped me. I differentiated the position function to get velocity, then found the extrema and checked before and after them. But I’m not getting choice E, like it says, and keep getting choice D. </p>
<p>For the second one, I can’t seem to get a negative sign before the square root, which is the right answer…</p>
<p>For the differential equation one, I’m getting this:</p>
<p>Y=(+/-)sqrt(2(2x^2-6))</p>
<p>dy/dx = 4x/y</p>
<p>ydy = 4xdx</p>
<p>y^2 /2 = 4x^2 /2 + C</p>
<p>at the P(2,-2), 2 = 8 + C, so C = -6</p>
<p>Then,</p>
<p>y^2 /2 = 4x^2 /2 - 6</p>
<p>y^2 = 4x^2 - 12</p>
<p>y = (+/-)sqrt(4x^2 - 12)</p>
<p>But when you substitute y back into dy/dx = 4x/y, at P(2,-2) it requires the negative sign to work out.</p>
<p>On the RHS, 4x/y equals 4(2)/(-2) = -4.</p>
<p>Speed is increasing when velocity and acceleration have the same sign.</p>
<p>These are helping a lot! Thanks :D</p>
<p>Hmmm, when I find the first and second derivatives, the extrema are x=5 and x=3, while the inflection is x=4. But after checking when the speed (velocity) is increasing, none of it works out to be choice E. I think it’s a typo 
</p>
<p>For clarification, to find when speed is increasing do you find when the acceleration is increasing or when the velocity is increasing?</p>
<p>There are two possible ways to find out when speed is increasing:
<p>Basically, you need to find out when the signs of v and a are the same.
You need to find both. Not just one.</p>
<p>So whenever Velocity and Acceleration are the same, the speed is increasing, even if they’re both negative? Now this is something I’ve never heard of O_O Thanks!</p>
<p>NP =)</p>
<p>If a and v are both negative, it is increasing in the negative direction (meaning it is moving faster to the left)</p>
<p>If a and v are both positive, it is increasing in the positive direction (meaning it is moving faster to the right)</p>
<p>hey, um, i’ve always been confused on how to check if a function is differentiable or not, could someone please help me?? :)</p>
<p>Check if it’s continuous at the point. If it’s not continuous at that point, then it’s definitely not differentiable.</p>
<p>Also, if it’s something like a piecewise function, the derivative from the left has to equal the derivative from the right.</p>