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We ended class today with an example similar to a problem on the turn in. To conclude the in-class example, we set up a function to find the distance between (2,0) and the curve as a function of . It is d(x)=sqrt((X-2)^2+x). We then want to find the critical values of this function by taking its derivative. For this, let's use a calculator to graph d'(x) and see where it is zero. We can then verify this value is a minimum by seeing that the graph of d'(x) is changing from negative to positive. We then should compare the value of d(x) here with the endpoint, d(0), which we know equals 2.
ReplyDeleteFor problem 4c on the turn in, should we use our answer from part b to find the rate of change of k?
ReplyDeleteYes
DeleteFor the book problem #23 on page 163, does anyone have hints for proving the particle is always moving right? Can I use the derivative? -Claire Westerlund
ReplyDeleteYes use the derivative. Think of which way the particle is moving when velocity is positive/negative.
DeleteThis comment has been removed by the author.
ReplyDeleteWhat are relative/local max and mins? Are they only relative/local max and mins when the absolute max/mins are not in the interval?
ReplyDelete-Sarah Mostofizadeh
Absolute maximums are the highest point that is hit from all of the domain values. Whereas the relative/local maximums are just high points within the domain. It all really depends on how you restrict it. At least that's how I've understood it from class!
DeleteRelative/local max and mins are the only ones we have when the absolute max and mins are not in the interval, yes, but they are also maximums and minimums in an interval that just aren't as massive as the absolute maxes and mins. Say there's a graph coming from -infinity that reaches a max, slopes downward to a minimum, and comes back up to a second, smaller maximum, before sloping back to negative infinity. If you're given an interval which includes both of these maximums, the first one will be the absolute maximum of the interval, the minimum (being the ONLY minimum) will be the absolute minimum, and the SECOND maximum will be a relative/local maximum. Hope this helps.
Delete-James Gruich
In 1a, I know it's similar to the in-class problem, but it asks for the other endpoint, as opposed to the minimum distance. I think the process is similar up to the point where you get the critical value of the distance formula's derivative. From there I'm having a hard time. Does the critical value have to be substituted as the x value in the distance formula to solve for the distance, and then the y value has to be solved for in the original distance formula? I've tried that but the distance ends up as 0, which makes no sense because they aren't touching. Any help would be appreciated.
ReplyDelete-James Gruich
This problem is identical to the in class example. You have a fixed point and a curve, and you are trying to find the point on the curve that is closest to the fixed point. A hint: you will want to solve the ellipse equation explicitly for y at some point...
Deletefor 1a, i found the distance between the drain and center of the ellipse, and the radius of the elipse that goes throught the drain point on the road, and subtracted them to find the distance from the road to the edge of the pond. is that correct? or am i completely wrong.
ReplyDeleteYou are assuming the line connecting the fixed point and he point on the curve closest to it also passes through the center of the ellipse. This may not be the case and therefore cannot be assumed.
DeleteWhen I solve solve for k in 4b I get an equation with x to the first. So when I go to take the derivative to answer 4c I get a constant value, does this mean my equation is wrong?
ReplyDelete-Cara Young
I get that you're trying to use the power rule to turn Xp to 1. However, when you take the derivative of Xp, you should instead turn Xp into (dXp/dt). Then you can substitute dXp/dt for 7.
Delete-Lizzy C, 1st hr