$ \frac{\operatorname dx}{\operatorname dt} = \cos (\frac{3}{2} t^2)$ ,$\frac{\operatorname dy}{\operatorname dt} = 3\sin(t^2)$; Tangent line? Q)An object moving along a curve in the xy-plane has position ( x(t), y(t) ) at time t, with dx/dt = cos (3/2 t^2) and dy/dt = 3sin(t^2) FOR 0 <= t <=3 at t=2, the object is at (4,5). write the equation for the tangent line at this point.
Comments:
First and for most what should i understand from (x(t), y (t)) I don't recall ever seeing a question set up this way.
I know very basic rules when it comes to finding a tangent line to a curve aka y=mx+b, take the derivative, plug in x and y values, etc. Upon reading this question I am completely baffled. I wish I had more to go on but my research online has been little to no help and I cannot find similar examples in my text book. Please help!
 A: How can we write the equation of a line? Usually in $\mathbb{R}^2$ we can write $y=mx+b$, but there is another way. A line is specified by giving two points. To turn this description into some way to generally describe lines, we need to think this way: these two points determine translation from a given origin plus the direction, given by the difference of these points. In other words, we can write that line as
$$\ell: (x_0,y_0) + t(p,q),$$
where $(x_0,y_0)$ is the component describing the translation and $(p,q)$ describes the direction.
It is usual to describe the motion of a partial with respect to a parameter, usually thought as time $t$. To do this, you have to describe the horizontal and vertical components of the particle as functions of $t$. In other words, you need to write $x=x(t)$ and $y=y(t)$.
In this exercise he has given you the derivatives of these functions. So what? Here's the catch: the tangent vector has its as components the time derivatives of the components functions of the motion. By plugging the time you want, you find the tangent vector at that point. This gives you the direction we have spoken of above.
It has given you the point where the particle starts, hence the translation information. Continue from here using this information.
A: For a parametric equation such as this you can find the slope of the tangent line by evaluating (dy/dt)/(dx/dt) at the time given.  Therefore, plug in t=2 into dy/dt and you get 3sin(4) and divide by dx/dt  which is equal to cos(6).  This will be the slope of the tangent line to the curve at t=2  then use the point given (4,5) to write the equation of the tangent line in your preferred form.
A: If you were wanting to look this kind of question up in a textbook, it would be under "parameterization" of curves or something like that. What's happened is they took a curve on the xy-plane and wrote in terms of some parameter t (which usually represents time in practical applications).
For your question:
To be clear, you are still working with a curve y = f(x). Only it's become slightly more complicated so it's more like: y(t) = f(x(t)).
First in a tangent line question you would need a point on your line. If we had y = f(x) then you're point could be written as (x,y) on the xy-plane.
The question tells you that the point you are looking at is $(4, 5)$. In other words, $x = 4$ and $y = 5$. And this occurs at "time" t=2.
Next you need the slope of your graph at this point. To find your derivative, you want $dy/dx$. Since you have $dy/dt$ and $dx/dt$, you can take $dy/dt \over dx/dt$ to get $dy\over dx$ (in an imprecise way, you can think of it as the $dt$'s cancelling out, like they would if you were dividing two fractions with the same denominator.)
When you calculate $dy/dx$ you will end up with some function in terms of $t$. So what you do is plug in the "time" they give you, $t=2$, into your derivative. This will give you a numerical value, $m$, for the slope, which will correspond to your point (4,5).
So now you have (x,y) = (4,5) and you have a value, $m$, for your slope. Just plug everything into the equation of a line to find your tangent line.
