Question

Find, correct to 2 decimal places, the angle between the tangents to y =3e^-x and y =2+ e^x at their point of intersection.

Answer 1

well, where do they meet at? that point might be good to know; and then we can derive these at with the point to find the tangents vector

Answer 2

2+e^x = 3e^(-x)

wolfram might help to find the intersection

Answer 3

(0,3), that was nice

Answer 4

soo, [3e^(-x)]' = -3e^(-x) at x=0; slope = -3

[2+e^x]' = e^x at x=0; slope = 1

Answer 5

the vectors are then <1,-3> and <1,1>, we can dot these to find the angle

Answer 6

1.1 + -3.1 -2 -2
-------------- = ------- = -------
sqrt(2) * sqrt(10) sqrt(20) 2sqrt(5)


cos(a) = -1/sqrt(5); a = cos-1(-1/sqrt(5)) if I did it right

Answer 7

116.57 maybe?

Answer 8

or; one angle is 45

the other is: tan(a) = -3

a = tan-1(-3) = |-71.565....|

a+45 = 116.565.....

Answer 9

same get up

Answer 10

the answer is 63.43 I dont know how they got it

Answer 11

http://www.wolframalpha.com/input/?i=y%3D2%2Be^x+and+y%3D3e^%28-x%29

this the graph so I know the point is good; I know the derives are good, so it must be in how im interpreting the angles

Answer 12

i got that far also

Answer 13

slope of 1 = 45 degrees; maybe they want the radians?

Answer 14

no matter how i do it I only see an angle of 116.57 degrees; you sure your answer is to this problem?

Answer 15

180 - 116.57 = 6.43

Answer 16

they wanted the supplement angle to the one we found

Answer 17

seeing how there are 2 angles formed by lines that cross; one of ours was 116.57 and the other was 63.43

Answer 18

ok cool

Answer 19

how they expect you to decide is beyond me lol

Answer 20

its assumed to be the acute angle

Answer 21

the way I learnt this way by memorising a formula\[\tan(\theta) = |\frac{m1-m2}{1+(m1)(m2)}|\]

Answer 22

where m1 and m2 are the gradients of the curves at the point of interest , I did use vectors or anything like that.

The absolute value in the formula gives a acute angle for the answer