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I'm confused about problem #5 on the homework from page 172. You have to find the derivative of y=e^(1+lnx) and the back of the book says the answer is e. For some reason I keep getting e^(1+lnx)(1/x). Where am I messing up?
ReplyDeleteOops, #15.
DeleteTry using exponent properties to rewrite y as y=e*e^lnx first.
DeleteIn problem 3, how do you know whether the critical point of acceleration is a minimum or a maximum point?
ReplyDelete-Andrew Saad
Plug in values for t within the interval but surrounding the critical point t value. By getting those values, you can determine if that critical point value is a max or min by comparing it to those values you just found.
Delete-Evan Gilman
Plug in values for t within the interval but surrounding the critical point t value. By getting those values, you can determine if that critical point value is a max or min by comparing it to those values you just found.
ReplyDelete-Evan Gilman
^ I meant for that to be a reply to Andrew's question.
DeleteIn problem 4d for the area equation how do you relate Xp and x?
ReplyDelete-Evan Gilman
This is a poorly written part of the question, as they never really define point Q for us other than in the diagram. We shall assume that Q is directly below P, thus P and Q share an x-coordinate.
DeleteYou can substitute the Xp in for the X value because the X value of Q is the X value of P and Xp is the x value of P.
ReplyDelete- Andrew Saad
How do you do #11 on the derivative review 1 worksheet? I don't even know where to start.
ReplyDeleteI started by changing the x/y and y/x fractions to xy^-1 and yx^-1 (product rule is easier to work with in this situation). Then take the derivative of the left side (product rule and chain rule) and the derivative of the right side (e^x rule and chain rule and product rule). Then distribute in, isolate dy/dx onto one side, factor it out, and divide it to the other side and simply if you want.
Delete- Jacob Edelson
*simplify
DeleteOn the turn in P.172 #23, what is the first step I should take when trying to find the derivative?
ReplyDeletee^tan^-1x has a constant in its base so you can use derivative of a^x=(a^x)(lna) and from there you would use product and chain rule
Delete-Michael Pastoria
Michael is correct...keep in mind the derivative rule for e^x to make life a bit simpler.
DeleteOn problem 1c on the turn in should i assume the dx/dt is still pi like it stated in 1b? Also, just to make sure I am on the right track, are we supposed to use the distance formula?
ReplyDeleteThe answer is "YES" to both questions you pose.
DeleteShould 1c involve the distance formula or is there something else i'm missing?
ReplyDeleteNevermind, I see that question was answered already!
DeleteIs plugging in numbers for #50 on page 424 enough proof of the limits?
ReplyDeleteFor problem 1a on the turn in can you assume that the line normal to the point on the curve will produce the smallest distance? If not, do you need to solve the original equation for x and plug it into the distance formula?
ReplyDeleteIt is not in fact true that the shortest distance lies on a perpendicular line. You must use the distance formula.
DeleteHow do I figure out the limit of arctan(t) as t approaches infinity? Do I need to make it applicable for L'Hopital's rule?
ReplyDeleteThis is for #23c on the bookwork.
Dahvi Lupovitch
Think of where the horizontal asymptotes are on the arctan(x) graph.
DeleteThis is a well known limit that you should commit to memory. In addition to Matt's suggestion, consider where tanx has vertical asymptotes.
DeleteFor the concavity worksheet, I'm still not sure how to determine when the graph increases or decreases just by the critical points. Any tips?
ReplyDeleteAfter you find the critical points, plug them into the second derivative to see whether the point is a minimum (output greater than zero) or maximum (output less than zero). If the critical point is a minimum, the derivative will be negative on the left side - decreasing, and positive on the right side - increasing. This will be true until another critical point is reached. Is that what you were asking? Matt Bachand
DeleteFor problem 4b on the turn-in, I know that we have to first find the slope of the equation in terms of Xp, but I'm confused as to what step I should take next. How do I get the x-value?
ReplyDelete-Emma Lucken
After you find the slope of the line in terms of Xp, the next step should be to use xp for every spot which you solved for xp in the first part. Since K is an x value on the tangent line, you can use that slope and the y, in terms of xp, to create an equation for the tangent line. Then solve for K. Was this what you were asking? Matt Bachand
DeleteSorry, didn't see that Jessie responded
DeleteYou need to use xp in place of everywhere you solved for xp in part a. You're solving for K as an x value on the tangent line, so you use that slope and the y in terms of xp to form a tangent line equation and solve for K from there. Does that help?
ReplyDeleteThis was supposed to be a reply to Emma's question not its own comment
Delete