How to find the normal to a curve at a specific point?

Question

Can someone walk me through how to find the normal to a curve at a specific point? Taking one example from my Calculus Textbook, it states Find the equation of the normal to the curve $sqrt(y)+xy^2=5$ at the point (4,1)?

Now, I know a Normal is perpendicular to a tangent line, and has a reverse-signed reciprocal slope to our tangent. But beyond the general knowledge there, how do I go about solving this as asked?

Answer 1

There are two basic approaches here:

1. Rewrite the defining equation for your curve to make one coordinate a function of the other. In this case it is easy to reach $$ x = \frac{5-\sqrt y}{y^2}$$ which you can differentiate symbolically to find a tangent vector and then produce the equation for the normal (goes through the specified point, and perpendicular to the tangent).

2. Know that the normal to a level curve $f(x,y)=c$ is always in the direction of the gradient of $f$, so compute $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ at $(4,1)$ construct a line through $(4,1)$ in that direction.

Answer 2

Write the curve in the form $f(x,y)=0$. Then the equation of the tangent line at $(x_0,y_0)$ is $$\partial_xf(x_0,y_0)(x-x_0)+\partial_yf(x_0,y_0)(y-y_0)=0.$$

Answer 3

We call $ \mathbf {r} $ the curve. To find the normal vector in a point $ p $ of the curve you derive the vector$ \mathbf {r'}/||\mathbf {r'}||$ and calculate in $ p $ the derivate. In your case first find a parametrization of the curve.

Answer 4

Hints: Implicit differentiation will give you an expression for the slope $\frac{dy}{dx}$ of the tangent in terms of the coordinates $x$ and $y$. Evaluating this expression at the point of interest will tell you the slope of the tangent line there, namely, $m_{\textrm{tangent}} = \left. \frac{dy}{dx}\right|_{(x,y)=(4,1)}$.

Then, you know the slope of the normal would be $m_{\textrm{normal}}=-1/m_{\textrm{tangent}}$.

At this point, you have the slope of the normal and a point on the normal, so you should be able to write down an equation of the normal.

You should be able to handle the case of a horizontal (or vertical) tangent pretty easily since the normal would then be vertical (or horizontal) and have an especially simple equation.