Question

Represent the plane curve by a vectorvalued function. (There are many correct answers.) $$ \frac{x^{2}}{16}-\frac{y^{2}}{4}=1 $$

Short answer

The vector-valued function of the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\) is \(\vec{r}(t)=4\sec(t)\vec{i}+2\tan(t)\vec{j}\).

Step-by-step solution

Identify the hyperbola's parameters

By looking at the equation \(\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\), we can identify the constants associated with \(x^{2}\) and \(y^{2}\). Here, these constants are 16 and 4 which represent \(a^2\) and \(b^2\), respectively, in the general equation of a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\). Thus, our parameters are \(a=4\) and \(b=2\).

Use the parameterized form of the hyperbola

Equations of hyperbolas can be parameterized. For this case, where the x-term is positive, the parameterization can be done like so \(x=a\sec(t)\) and \(y=b\tan(t)\) where \(t\) is a parameter.

Apply the parameterization

Apply the parameter values obtained from step 1 into the parameterization formula from step 2. This gives the parametric equations \(x=4\sec(t)\) and \(y=2\tan(t)\).

Write as vector-valued function

Compile these equations into a single vector-valued function. We can assign the x-coordinates to component \(i\) and the y-coordinates to component \(j\). Therefore, the vector-valued function will be \(\vec{r}(t)=4\sec(t)\vec{i}+2\tan(t)\vec{j}\).