# Thread: Several Problems

1. ## Several Problems

I've got these three problems that I just can't work out. If someone could explain one or all of them, that would be fantastic

1) Find a formula for the Y(Nth derivative) for X^4+Y^4=A^4

^^^(Derivative of X with respect to Y)^^^

2) Suppose the curve X^4+aX^3+bX^2+cX+d has a tangent line with a formula y=2x+1 at x=0, and one with a formula y=2-3x at x=1. Find the values of a, b, c, and d.

3)Suppose that XY=C has a tangent line at point P.

a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P.

b) Show that the triangle formed always has the same area, no matter where P is located on the graph of the function.

You guys have yet to let me down, and I'm really counting on you this time! Thanks!

2. Okay, here's what I got for #2...

$\displaystyle f(x)=x^4+ax^3+bx^2+cx+d$

$\displaystyle \rightarrow f'(x)=4x^3+3ax^2+2bx+c$

Now we can tell a few things from the information that they give us from the tangent lines. If the tangent at $\displaystyle x=0$ is $\displaystyle y=2x+1$, then we know that $\displaystyle f'(0)=2$ because the derivative takes the value of the slope of the tangent line. We also know that $\displaystyle f(0)=1$ because $\displaystyle y=2x+1$ is tangent at $\displaystyle x=0$ and $\displaystyle 2(0)+1=1$. Likewise the other tangent line tells us that $\displaystyle f'(1)=-3$ and $\displaystyle f(1)=-1$ with this information we can start chipping away at the coefficients. Let's start with the derivative:

$\displaystyle f'(0)=2$
$\displaystyle \Rightarrow 4(0)^3+3a(0)^2+2b(0)+c=2$
$\displaystyle \Rightarrow c=2$

using this and our other derivative point we find...

$\displaystyle f'(1)=-3$
$\displaystyle \Rightarrow 4+3a+2b+2=-3$
$\displaystyle \Rightarrow 3a+2b=-9$

this equation of a and b may seem like a dead end, but we'll use it later. Now let's look at the function itself and the points that we have for it. notice that I'm going to continue using c=2.

$\displaystyle f(0)=1$
$\displaystyle \Rightarrow (0)^4+a(0)^3+b(0)^2+2(0)+d=1$
$\displaystyle \Rightarrow d=1$

using this information and the other function point we discover...

$\displaystyle f(1)=-1$
$\displaystyle \Rightarrow (1)^4+a(1)^3+b(1)^2+2(1)+1=-1$
$\displaystyle \Rightarrow a+b=-5$

lo and behold, we have yet another equation of a and b. Now we can set up a linear system to solve for the last two coefficients:

$\displaystyle 3a+2b=-9$
$\displaystyle a+b=-5$

you can solve this by whatever method suits you, and you'll discover that a=1 and b=-6.

collecting our results:
a=1
b=-6
c=2
d=1

3. ## #3

In order to answer the first part of this question, we need to figure out how we can find the midpoint of the line segment in question. In order to do this we need to notice that the endpoints of the line segment in question will be the x-intercept and y-intercept of the associated tangent line. So we need to find an equation for the line tangent to $\displaystyle xy=c$.

in order to avoid having to do implicit differentiation, I'm going to rewrite $\displaystyle xy=c$ as $\displaystyle y=cx^{-1}$. This means that $\displaystyle y'(x)=-\frac{c}{x^2}$

Now we need to write the equation of the line tangent to $\displaystyle y=cx^{-1}$ at some arbitrary point $\displaystyle (x_0, y_0)$. Using the point-slope formula we get:

$\displaystyle y-y_0=m(x-x_0)$

$\displaystyle \rightarrow y-y_0=y'(x_0)(x-x_0)$

$\displaystyle \rightarrow y-y_0=-\frac{c}{x_0^2}(x-x_0)$

now that we have an equation of the line tangent to $\displaystyle (x_0, y_0)$, we can find the intercepts. Lets start with the x-intercept, which occurs when y=0...

$\displaystyle y=0$

$\displaystyle \rightarrow -y_0=-\frac{c}{x_0^2}(x-x_0)$

$\displaystyle \rightarrow -y_0=-\frac{cx}{x_0^2}+\frac{c}{x_0}$

$\displaystyle \rightarrow x=\left(y_0+\frac{c}{x_0}\right)\left(\frac{x_0^2} {c}\right)$

$\displaystyle \rightarrow x=\frac{x_0^2y_0}{c}+x_0$

$\displaystyle \rightarrow x=\frac{x_0x_0y_0}{c}+x_0=x_0+x_0=2x_0$

$\displaystyle \therefore xint\ at\ (2x_0,0)$

if you're wondering about the second to last line, remember that since $\displaystyle (x_0,y_0)$ is a specific point on the curve $\displaystyle xy=c$, it is also true that $\displaystyle x_0y_0=c$. By a similar process, we can set x=0 and discover that the y-intercept occurs at $\displaystyle (0,2y_0)$.

now we apply the midpoint formula to the two intercepts that we have found--($\displaystyle (2x_0,0)$ and $\displaystyle (0,2y_0)$)...

$\displaystyle \left(\frac{2x_0+0}{2},\frac{0+2y_0}{2}\right)=(x_ 0,y_0)$

so we have shown that the midpoint of any line segment formed in the prescribed manner will be the point tangent to $\displaystyle xy=c$.

as for the area of the triangle formed from the points $\displaystyle (2x_0,0)$, $\displaystyle (0,2y_0)$, and $\displaystyle (0,0)$, simply note that the triangle has a base of $\displaystyle 2x_0$ and a height of $\displaystyle 2y_0$, and apply the formula for area of a triangle...

$\displaystyle A=\frac{1}{2}bh=\frac{1}{2}(2x_0)(2y_0)=2x_0y_0=2c$

and so we see that the area of the triangle will always be 2c regardless of our choice of $\displaystyle x_0$ and $\displaystyle y_0$

4. Sorry, I've got no idea about how to do #1.