# Differentiation of exponentials help plz:) (1 Viewer)

#### csi

##### New Member
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

##### Well-Known Member
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 ($\bg_white \phi$) is given by

$\bg_white \tan\phi=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$

where $\bg_white m_1$ and $\bg_white m_2$ are the gradients of the tangents to each curve at P.

#### beetree1

##### Active Member
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

##### New Member
If two curves meet at a point P, the acute angle between them ($\bg_white \phi$) is given by

$\bg_white \tan\phi=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$

where $\bg_white m_1$ and $\bg_white m_2$ are the gradients of the tangents to each curve at P.
thanks

#### CM_Tutor

##### Well-Known Member
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
$\bg_white \frac{d}{dx}e^x = e^x \implies \frac{d}{dx}e^{f(x)}=e^{f(x)} \times f'(x) \qquad \text{by the Chain Rule}$

\bg_white \begin{align*} \text{So, if}\quad y&=e^{-kx} \\ \text{Then,}\quad \frac{dy}{dx}&=e^{-kx} \times \frac{d}{dx}(-kx) =e^{-kx} \times -k = -ke^{-kx} \\ \text{And,}\quad \frac{d^2y}{dx^2}&=-k \times \frac{d}{dx}e^{-kx} =-k \times e^{-kx} \times \frac{d}{dx}(-kx) = -ke^{-kx} \times -k = k^2e^{-kx} \end{align*}

Using these results an substituting into the given differential equation, you should get that $\bg_white e^{-kx}\left(k^2-4k+3\right)=0$. The rest should follow so long as you remember how an exponential graph appears, and thus realise that $\bg_white e^{-kx}>0$ for all $\bg_white k\in\mathbb{R}\neq0$ and for all $\bg_white x\in\mathbb{R}$.

#### csi

##### New Member
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

##### Well-Known Member
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...

$\bg_white y = x^2 - 4x + 2$

$\bg_white \text{At x = 0:} \quad y = 0^2 - 4(0) + 2 = 0 - 0 + 2 = 2$

So, the point $\bg_white P(0,\ 2)$ lies on $\bg_white y=x^2-4x+2$.

#### TheShy

##### Member
$image=https://latex.codecogs.com/png.latex?\bg_white+\tan\phi=\left|\frac{m_1-m_2}{1+m_1m_2}\right|&hash=2aa8802158b3c0d85b21166ef13e9490$
Is this still in the syllabus? I thought they removed it

#### tito981

##### Member
Is this still in the syllabus? I thought they removed it
i think they removed the use of perpendicular distance formula.

#### CM_Tutor

##### Well-Known Member
Is this still in the syllabus? I thought they removed it
They certainly kept $\bg_white m=\tan\theta$ 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

##### Well-Known Member
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

##### Member
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

##### Well-Known Member
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

##### Active Member
Yes, it does...

$\bg_white y = x^2 - 4x + 2$

$\bg_white \text{At x = 0:} \quad y = 0^2 - 4(0) + 2 = 0 - 0 + 2 = 2$

So, the point $\bg_white P(0,\ 2)$ lies on $\bg_white y=x^2-4x+2$.