I cannot solve this problem: Find the normal to the function: $$ x(t) = \ln(t^2 +1)$$ $$ y(t) = \ln(t^4 + 1)$$ which is parallel to $ y = \frac{1}{2}x + \pi$.
Can anybody solve it step by step?
I know how to find the points $(t)$ in which this funtion has to have a normal. By my calculations its $-1$ and $1$. What should I do next?
$\frac{dx}{dt} = \frac{2t}{t^2 +1}$
$\frac{dy}{dt} = \frac{3t^3}{t^4 + 1}$
Tangent = $\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx}$
Normal = -$\frac{dx}{dy}$
Other equation is of the form
$$y = mx + c$$
where $m$ is tangent.
Find normal then compare both to get the values of $t$ as both are parallel.
Such $t$ should verify: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\ldots =\frac{2t^2(t^2+1)}{t^4+1}=-2.$$ We get $2t^4+t^2+1=0,$ and the biquadratic equation has no real roots so, there is no $t$ satisfying the given conditions.
We need to find the point(s) where the normal vector is parallel to the vector $\langle 2, 1 \rangle$. That could only occur when the tangent at that point is normal to the vector $\langle 2,1 \rangle $ because the normal and tangent make $90$ degrees.
We have $r(t)=\langle \ln (t^2+1), \ln (t^4+1) \rangle$.
So ,taking the derivative, the tangents are given by:
$$\langle \frac{2t}{t^2+1}, \frac{4t^3}{t^4+1} \rangle$$
We need the dot product of this and $\langle 2,1 \rangle $ to be $0$. That is only possible when,
$$2(2t)+1(4t^3)=0$$
$$t(t^2+1)=0$$
So $t=0$ is gives the only possible tangent that may work. This gives tangent $<2,4>$ and hence a normal vector $c<-4,2>$ which is clearly not parallel to $<2,1>$. So there is no such normal.