A circle centered at the origin is tangent to $y=2^x$. What is the radius of the circle?

Question

I feel as though I am doing the analytical part correctly, however, where I am facing the roadblocks in this problem is in the actual algebra itself. Perhaps I am not doing something right in my calculus to begin with, but here is what I have done so far in attempting to answer this problem:

If the two curves defined by $x^2 + y^2 = r^2$ and $y=2^x$ are tangent to each other, then the following conditions must be true: the curves themselves are equivalent at the point of tangency, and their derivatives are also equivalent at the point of tangency. Describing this mathematically: $2^x = \sqrt{r^2-x^2}$

$2^xln2$ = $\frac {-x}{r^2-x^2}$

I solved the first equation for $r^2$:

$2^{2x} = r^2-x^2$

$2^{2x} + {x^2} = r^2$

Substituting this back into the second equation:

$2^xln2$ = $\frac {-x}{\sqrt{2^{2x}}}$

$2^xln2$ = $\frac {-x}{2^x}$

This poses an equation that, after many attempts and various different methods, I cannot seem to solve algebraically. I know that there is a solution because I could find one using an online grapher, but my calculus teacher expects an algebraic arrival at the answer. My plan was to find $x$ and then use $x$ to find $r$ using the relationship between them established in one of the equations.

Any help is greatly appreciated!

Answer 1

The distance of the point $0=(0,0)$ to $P=(x,2^x)$ is $$d(0,P)=\sqrt{x^2+2^{2x}}$$ It is enough to find the minimum of this distance because the distance of the origin to other point of the curve would be greater than the radius of the circle tangent to the curve. The derivative, $$\frac{2x+2\cdot2^x\space\ln x}{\sqrt{x^2+2^{2x}}}=0$$ gives $$x\approx -0399$$ so the minimum is equal to $$\sqrt{(-0.399)^2+2^{-0.798}}\approx \color{red}{ 0.85694\approx r}$$

Answer 2

You are going to need the Lambert $W$-function. The Lambert function cannot be represented in terms of the elementary functions, unfortunately.

Your equation is:

$$-\frac{x}{2^x} = \log(2) 2^x$$

or $$-x4^{-x} = \log(2)$$.

Multiplying both sides by $\log(4)$, you get an equation:

$$(-x\log 4)e^{-x\log 4} = \log(2)\log(4)$$

This is of the form $ze^z=C$, which is what the Lambert $w$-function solves. Letting $z=-x\log 4$, you have:

$$ze^z = \log(2)\log(4)$$ So $z=W(\log(2)\log(4))$, and $$x=-\frac{z}{\log 4} = -\frac{W(\log(2)\log(4))}{\log(4)}$$

I get $x\approx -0.39878$. Then $y\approx 0.758498$. Then we get:

$$\begin{align}y\log(2)&\approx 0.525751\\ \frac{-x}{y}&\approx 0.525750\end{align}$$

Not bad, given the limits of the Lambert calculator I used.

Now compute $\sqrt{x^2+y^2}$.

Note that $y^2=4^x=e^{-z}=\frac{z}{\log(2)\log(4)}=-\frac{x}{\log 2}$. Not sure how that will give us a better estimate of the radius. It still doesn't let you evaluate that without a numerical calculation. Still, given the numerical expression for $x$, the radius is:

$$r=\sqrt{x^2-\frac{x}{\log 2}}$$

Noting that $x=\frac{-z}{2\log 2}$, we see that we have:

$$r=\frac{1}{2\log 2}\sqrt{z^2+2z}$$