calculusparametric
Consider the curve defined by the parametric equations $x=t^2 +t-1$ and $y=te^{2t}$
i) Show that $dy/dx =e^{2t}$
ii) Hence show that the tangent to the curve at the point on the curve where $t= -1$ passes through the origin.
I'm sorry to bug you guys, but I'm clueless and would help me if someone could help me, so I get questions that are similar to this. Thanks!
take a hard look at picture take some fixed point anywhere on the circle and when the circle is rolled around the curve drawn by the the fixed point is the TROCHOID
in the diagram i have drawn i labelled it as point D some where on circle also i have taken the angle to be "a"
from the diagram you can see that point D is online CP and P started initially from origin and it has travelled by an angle of "a" so PR=OR= ra ( PCR is a sector with angle a and arc length ra)
our job is to find the x and y co-ordinates of D which is making the curve
from triangle CDQ
CD=d which is given in our question
CQ= d $cosa$ and DQ= d $sina$
now x-coordinate of point D= OR- DQ
SO x= r*a-d $sina$
and y- coordinate of point D = CR- QR
y=r- d $cosa$
hence the parametric representation of the curve is
x= r*a-d $sina$
y=r- d $cosa$
In $\Re^3$, a plane can be characterized completely by a vector normal to the plane, and a point it goes through:
Imagine any plane. Note that every vector normal to it's surface is necessarily parallel to each other (even if you consider the normal pointing in the other direction, the lines all these normals live in are parallel). So, once you have a line you have your plane except for a translation in the direction of that normal, so you need to specify a point in the space that belongs to the plane to fully define it.
Now, in equations, a vector connecting to points in the plane must be perpendicular to $\vec{v}=(3,1,-6)$. Let's call this generic point in the plane $\vec{u}=(u_x,u_y,u_z)$ and for a second let $\vec{u}_0$ be the point in the plane that they are giving us (origin in this case), then $$0=\vec{v}\cdot(\vec{u}-\vec{u}_0)=3u_x+u_y-6u_z$$
Please, note that what needs to be perpendicular to $\vec{v}$ is not a point in the plane $\vec{u}$ but a vector contained (or parallel) to the plane itself and therefore we need to use $(\vec{u}-\vec{u}_0)$ in the dot product above. Since the origin is contained in the plane in your case, then the distinction is hard to grasp but it is very important to understand.
Best Answer
$$\frac{dy}{dt}=e^{2t}+2te^{2t}=e^{2t}(1+2t)$$
$$\frac{dx}{dt}=2t+1 $$
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
At $t=-1,x=-1, y=-e^{-2}$
So, the equation of the tangent will be $$\frac{y-(-e^{-2})}{x-(-1)}=\frac{dy}{dx}_{\text{(at }t=-1)}=e^{-2}\implies y=x\cdot e^{-2}$$ which clearly passes through the origin $(0,0)$