Differentiation of exponentials help plz:) (1 Viewer)

csi

Member
Joined
Nov 10, 2019
Messages
94
Gender
Undisclosed
HSC
2021
Hi, I’m kinda stuck on these question from my hw:

1. the curves y=x^2-4x+2 and y=e^x +1 intersect at the point (0,2) find the acute angle to the nearest degree between the two curves at this point.
2. the function y=e^-kx satisfies d^2y/dx^2+4(dy/dx)+3y=0.
(a) show that k^2-4k+3=0
(b) hence find the possible values of k

Much appreciated if anyone could help :)) thanks guys
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
1. the curves y=x^2-4x+2 and y=e^x +1 intersect at the point (0,2) find the acute angle to the nearest degree between the two curves at this point.
If two curves meet at a point P, the acute angle between them () is given by



where and are the gradients of the tangents to each curve at P.
 

beetree1

Well-Known Member
Joined
Feb 10, 2019
Messages
539
Gender
Female
HSC
2020
this might just be me but is there a typo in q1 because i dont think y=x^2-4x+2 even has a point (0,2)?
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
2. the function y=e^-kx satisfies d^2y/dx^2+4(dy/dx)+3y=0.
(a) show that k^2-4k+3=0
(b) hence find the possible values of k




Using these results an substituting into the given differential equation, you should get that . The rest should follow so long as you remember how an exponential graph appears, and thus realise that for all and for all .
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
Is this still in the syllabus? I thought they removed it
They certainly kept and the double-angle formulae, so it can be derived in two lines.

Also, remember that knowing what is truly covered depends on both the syllabus and HSC exams, and we only have the former.

For example, the old topic "locus and the parabola" is supposed to have been removed, but about a half of it is still within the syllabus as written.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
i think they removed the use of perpendicular distance formula.
But that can be worked around by using vectors... find a vector v along the required line, then a vector u through the point and make the dot product zero, etc.
 

tito981

Well-Known Member
Joined
Apr 28, 2020
Messages
326
Location
Orange
Gender
Male
HSC
2021
But that can be worked around by using vectors... find a vector v along the required line, then a vector u through the point and make the dot product zero, etc.
iirc perpendicular distance was in the old 2u course, so would the use of vector dot products and its link to perpendicular distance be only tested in 3u?
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
iirc perpendicular distance was in the old 2u course, so would the use of vector dot products and its link to perpendicular distance be only tested in 3u?
Good point - clearly the use of vectors would make such a question impossible below MX1 level
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top