How do you find the horizontal and vertical tangents to x = cos(3t) and y = 2sin(t)?

Hint: We will first find the derivative of x with respect to t and then equate that equal to 0 to find the vertical tangent, now we will find the derivative of y with respect to t and equate it equal to 0 to find the horizontal tangent.

Complete step by step answer:
We are given that we are required to find the horizontal and vertical tangents to x = cos (3t) and y = 2 sin(t).
Consider x = cos (3t) for once.
Taking the differentiation with respect to t on both the sides of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{d}{{dt}}(x) = \dfrac{d}{{dt}}\left( {\cos 3t} \right)$
Simplifying both the sides of the above equation, we will then obtain the following equation:-
$ \Rightarrow \dfrac{{dx}}{{dt}} = - 3\sin 3t$
Now, equating it equal to 0, we will obtain the vertical tangent.
Therefore, the vertical tangent is given by: - 3 sin 3t = 0
Therefore, the possible values of 3t are 0 or $n\pi $ which implies that the possible values of t are 0 or $\dfrac{{n\pi }}{3}$, where n is an integer.
Hence, the vertical tangents are $t = 0,\dfrac{{n\pi }}{3}$.
Consider y = 2 sin(t) now.
Taking the differentiation with respect to t on both the sides of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{d}{{dt}}(y) = \dfrac{d}{{dt}}\left( {2\sin t} \right)$
Simplifying both the sides of the above equation, we will then obtain the following equation:-
$ \Rightarrow \dfrac{{dy}}{{dt}} = 2\cos t$
Now, equating it equal to 0, we will obtain the horizontal tangent.
Therefore, the horizontal tangent is given by: 2 cost (t) = 0
Therefore, the possible values of t are $n\pi + \dfrac{\pi }{2}$ , where n is an integer.

Hence, the horizontal tangents are $t = n\pi + \dfrac{\pi }{2}$.

Note: The students must notice that the cosine of an angle takes the value 0 when angle is of the form $n\pi + \dfrac{\pi }{2}$ and the sine of any angle is zero when it is of the form 0 or $n\pi $.
Therefore, when we equate 3t to be equal to 0 or $n\pi $, therefore, we obtained the value of t as 0 or $\dfrac{{n\pi }}{3}$, where n is an integer.
The students must also know that the form in which the equation of the curve is given to us is the parametric form of the equation which we used to find the vertical and the horizontal tangents.