Differentiation of exponentials help plz:) (1 Viewer)

[csi]

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]

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]

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)?

 

[csi]

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.

thanks

 

[CM_Tutor]

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 .

 

[csi]

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)?

dw i’ve got it

 

[CM_Tutor]

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)?

Yes, it does...





So, the point lies on .

 

[TheShy]

Is this still in the syllabus? I thought they removed it

 

[tito981]

Is this still in the syllabus? I thought they removed it

i think they removed the use of perpendicular distance formula.

 

[CM_Tutor]

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]

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]

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]

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

 

[beetree1]

Yes, it does...





So, the point lies on .

lol my bad