# Slope function of a curve

I have a question I am having trouble answering:

The slope function of a curve is: $$\frac{dy}{dx}=ux+k$$

u and k are constants.

The curve passes through $(0,-1)$ and $(2,-5)$

At $(2,-5)$ the slope equals $1$.

How do I get the equation?

I have tried integrating and pluging the two above points in and trying to solve for u and k but I end up getting the integration constant $c=-1$ and $k=2-u$ I then plug this into the integrated slope function: $$\frac{ux^2}{2} + kx -1$$

but I don't get the correct answer.

Could someone point out where I am going wrong wrong and how to correct it.

• you meant you first "equation" to read $\frac{dy}{dx} = ux + k$, right? – Robert Lewis Sep 3 '13 at 22:00
• Yes, sorry I did – user Sep 3 '13 at 22:05
• It looks like you solved for $c$ correctly, but I imagine you made an algebra mistake somewhere on the way to solving for $k=-5$ and $u=3$. – parsiad Sep 3 '13 at 22:05

With

$y' = ux + k$

we have in general

$y = \frac{1}{2}ux^2 + kx + c$,

for some constant $c$. Plugging in the point $(0, -1)$ yields

$c = -1$;

plugging in $(2, -5)$ gives

$2(u + k) = -4$

and, last but not least, plugging $x = 2$, $y' = 1$ into the slope equation gives

$2u + k = 1$.

Then

$2k= -4 - 2u = -4 - (1 - k)$,

whence

$k = -5$;

and

$u = 3$

immediately follows.

Hope this helps. Cheers.

• Thanks, This is perfect, I see my mistake – user Sep 3 '13 at 22:52
• @user2352274: Glad to be of service! – Robert Lewis Sep 3 '13 at 23:06

From $\frac{dy}{dx}=ux+k$ you have $y=\frac{ux^{2}}{2}+kx+c$. The three pieces of information yield the equations \begin{align*} -1 & =0u+0k+1c\\ -5 & =2u+2k+1c\\ 1 & =2u+1k+0c. \end{align*} You can describe this by the matrix equation $$M\mathbf{x}=\mathbf{b}$$ where $$M=\left[\begin{array}{ccc} 0 & 0 & 1\\ 2 & 2 & 1\\ 2 & 1 & 0 \end{array}\right]$$ and $$\mathbf{b}=\left[\begin{array}{c} -1\\ -5\\ +1 \end{array}\right].$$ Note that the way I've written it, $$\mathbf{x}=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ Solving, $$\mathbf{x}=M^{-1}\mathbf{b}=\left[\begin{array}{c} +3\\ -5\\ -1 \end{array}\right]=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ This is equivalent to going through and solving each equation in terms of one of the independent variables.

• Could you write it as individual equations please, it is just a bit clearer for me as I know what the answer should be. Thanks – user Sep 3 '13 at 22:06
You solved for $c$ correctly, but you made a mistake with $k$. Note that when you plug in the point $(2, -5)$, you should get: \begin{align*} \frac{u(2)^2}{2} + k(2) - 1 &= -5 \\ 2u+ 2k &= -4 \\ u+ k &= -2 \tag{1}\\ \end{align*}
Now since the slope at $(2,-5)$ is $1$, we also know that: \begin{align*} \frac{dy}{dx}&=ux+k \\ 1 &= u(2) + k \\ -2u - k &= -1 \tag{2}\\ \end{align*}
Adding together $(1)$ and $(2)$ yields: $$-u=-3 \iff u = 3$$ Hence, using $(1)$, we obtain $k = -2-u = -2-(3) = -5$.