Newton Raphson method to find intersection points

I'm supposed to use N-R method to find intersection point near $(4,-4)$ of $x^2+y^2=16$ and $e^{1/3x} + (1/5)y =1 .$

So, I equated these two functions, and got $f(x) = 5- (5*\exp(x/3)) + (16- x^2)^{1/2}$ and $f'(x)= (-5/3)*\exp(x/3) - (16-x^2)^{-1/2}$.

As stated in my assignment, I'm supposed to start with point $(4,-4)$ , but as you can see...this would give me $-\infty$ for $f'(x).$ So I can't even go on to the next iteration.

What am I doing wrong?

• I would square the exponential and equate $y^2$, so that your derivative doesn't get that square root in the denominator. – Adrian Keister Feb 28 '18 at 21:47
• I suspect you were intended to locate the point of intersection by finding a "root" of a system of two equations in two unknowns. Newton-Raphson is the name we usually call Newton's method for a multivariable system of equations. That said, it is possible to eliminate one variable and proceed by the single variable Newton's method. You may have to pick a different starting point to avoid a singularity. What kind of explanation would you prefer? – hardmath Feb 28 '18 at 21:50

Way 2: Work with $u+v=16,e^{\sqrt{u}/3}-\sqrt{v}/5=1$, which is effectively working with $(x^2,y^2)$ as your variables under the assumption $x>0,y<0$. Then you can eliminate one variable without triggering any singularity.
Way 3: Do what you said, but start somewhere else and hope you get the right intersection. For instance you could start at $(\sqrt{8},-\sqrt{8})$ which is on the circle (and not on the axes) and will thus remove the singularity. (This might have actually been the point where they wanted you to start, i.e. $(4,-4)$ might have just been a mistake.)
However, you idea of eliminating $y$ and just solve for $x$ was good and works.
You have two equations $$e^{x/3}+\frac{y}{5}-1=0 \tag 1$$ $$x^2+y^2-16=0 \tag 2$$ Frm $(1)$, $y=5-5 e^{x/3}$. Plug it in $(2)$ to get $$x^2-50 e^{x/3}+25 e^{2 x/3}+9=0$$ So, you look for the zero of $$f(x)=x^2-50 e^{x/3}+25 e^{2 x/3}+9$$ $$f'(x)=2 x-\frac{50 }{3}e^{x/3}+\frac{50}{3} e^{2 x/3}$$ Now, Newton iterates using $x_0=4$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.000000000 \\ 1 & 2.943256168 \\ 2 & 2.166300869 \\ 3 & 1.755176003 \\ 4 & 1.649488097 \\ 5 & 1.643297711 \\ 6 & 1.643277453 \end{array} \right)$$ which is the solution for ten significant figures.