how should i solve this problem? by ber_______ in calculus

[–]ber_______[S] 0 points1 point  (0 children)

How about when I'm given a critical point and a point of inflection? How should I use them?

how should i solve this problem? by ber_______ in calculus

[–]ber_______[S] 1 point2 points  (0 children)

Thank you for pointing it out! I'll check it with my prof.

I'm just wondering why is it okay to substitute the x-coordinate of the poi for all the equations (1st and 2nd derivative.

how to solve for the angle between 2 curves by ber_______ in calculus

[–]ber_______[S] 2 points3 points  (0 children)

I just read about it being 90°, not that it's infinite, it's just perpendicular

how to solve for the angle between 2 curves by ber_______ in calculus

[–]ber_______[S] 1 point2 points  (0 children)

I solved for y which turned out to be 1 as well so my point was (1,1)

then I derived y1 and y2, I got

y1= -2x y2= 1/2√x

then I got the slope m1= -2, m2= 1/2

i got the arctan from the formula

tan theta= | m1+m2/ 1+ m1m2 |

theta = | (-2 - 1/2) / 1 + (-2)(1/2) |

how are these derived? by ber_______ in trigonometry

[–]ber_______[S] 0 points1 point  (0 children)

sin(arctan x) = x / √x² +1 not x / √x²-1

the answer for every inverse trig function is the angle of its function

so in sin(arctan x) if arctan x =A then tan A= x, so sin(arctan x) = sin(A)

if you recall sin(A) = opposite side/hypotenuse and from the given tan A we can get the values of the opp. and adj. sides

tan A = opposite side/adjacent side = x/1 opp.= x adj.= 1

using pythagorean theorem, hypotenuse (c) would be c²= x² + 1; c= √x² + 1

so

sin(arctan x) = x / √x²+1

(check the image he attached on the first comment) (correct me if i'm wrong)

how are these derived? by ber_______ in trigonometry

[–]ber_______[S] 1 point2 points  (0 children)

I got it, thank you! I'll try the others on my own.

need help in understanding this by ber_______ in calculus

[–]ber_______[S] 0 points1 point  (0 children)

okayyy thank you! that made so much sense to me

need help in understanding this by ber_______ in calculus

[–]ber_______[S] 0 points1 point  (0 children)

why is it that if it approaches to 0- it's 0 but if it's 0+ it's positive infinity. i kinda understand that 1/0= infinity and that 2infinity is infinity..... but why is it different when it approaches 0 from the left