Recent content by NYmike

Sorry, but I am not sure where you are going with this. When d = \sqrt{x^2 + y^2} : \frac{dd}{dt} = \frac{1}{2} (x^2 + y^2)^\frac{-1}{2} (2x\frac{dx}{dt} + 2y\frac{dy}{dt}) \frac{dd}{dt} = \frac{1}{2} * \frac{1}{\sqrt{x^2+y^2}} * 2xa + 2yb \frac{dd}{dt} = \frac{xa +...

I was taught that \frac{d(some var)}{dt} = the Rate of Change of some var. with respect to time. The rate of change of x with respect to time is 'a', and of y with respect to time is 'b'. I am trying to find the rate of change of distance (var. d) with respect to time. I am pretty sure that...

:eek: Even after plugging in the deriv is ridiculous. There must be a way to plug in something else. Am I using the wrong coordinates/formula?

d = \sqrt{\Delta x^2 + \Delta y^2} \frac{dx}{dt} = a \frac{dy}{dt} = b That's what I've come up with so far. Now, since I am trying to find the rate of change of the distance, then I believe I am looking for \frac{dd}{dt} Taking the derative of the distance formula directly would...

Hi. I am drawing a complete blank on this calc problem. Point a moves along the x-axis at the constant rate of 'a' ft/sec while point b moves along the y-axis at the constant rate of 'b' ft/sec. Find how fast the distance between them is changing when A is at the point (x,0)and B is at...

Hello. I am stuck on this physics problem that i have for homework. Its the wording of the question that's odd to me. I am not sure exactly where to begin. So after one century, the length of each day will be 24 hours and .001 s? And after the second century the length of each day will be...

pssh...I was thikning way too into it...thanks for your help. I used the formula vf^2 = vi^2 + 2ad Mathematically the answer turns out to be a positive number...Since it is still in the downward direction however, i should make it negative, correct? EDIT: And thanks for moving the...

Hello all. I am currently taking an AP Physics class in high school. My teacher has given us a homework assignment involving projectile motion, however, it is different than anything else we have done. Usually, the angle is upward, or directly horizontal...This question is downward. My...