Cartesian equation question

Dec 2014
31
0
Australia
Need help with a question:

Point p is moving on a curve given by:

x=bt^2, y=2bt

now the Cartesian equation i worked out to be is:

y=2b(x/b)^0.5 (as you make t the subject of x)

The part i am stuck on is finding the equation of the normal at P (bt^2,2bt), which is x+y=at^3+2at i think but not to sure how to show, if Q was the intersection of the normal with the x axis i let y=0 and i know how to do that, how to calculate the velocity at Q when t=7 seconds?

i have figured out this as well:

dx/dt= 2bt, dy/dx= 2b so dy/dx=1/t?

thanks giovanni
 
Feb 2015
2,255
510
Ottawa Ontario
x=bt^2, y=2bt

now the Cartesian equation i worked out to be is:

y=2b(x/b)^0.5 (as you make t the subject of x)
HOW did you arrive at that?

x=bt^2 : t = SQRT(x/b)

y=2bt : t = y / (2b)
So:
y / (2b) = SQRT(x / b)
y^2 / (4b^2) = x / b
y^2 = 4bx
y = 2SQRT(bx)
 
Sep 2015
11
2
Seoul
There is a curve defined as:

\(\displaystyle x=bt^2, y=2bt\)

(I'm going to assume \(\displaystyle b\neq 0\))

Since \(\displaystyle \frac{dx}{dt} = 2bt\), \(\displaystyle \frac{dy}{dt} = 2b\), we can say that the tangent vector at point \(\displaystyle P (b{t_{0}}^2, 2bt_{0})\) (which is when \(\displaystyle t=t_{0}\)) is \(\displaystyle (2bt_{0}, 2b)\).

So the equation for the normal line would be:

\(\displaystyle (x-b{t_{0}}^2, y-2bt_{0}) \cdot (2bt_{0}, 2b) = 0\)

\(\displaystyle 2bt_{0}(x-b{t_{0}}^2) + 2b(y-2bt_{0})=0\)

\(\displaystyle y=-t_{0}x+b{t_{0}}^3+2bt_{0}\)


Velocity is defined as: \(\displaystyle \overrightarrow{v}=\left( \frac{dx}{dt}, \frac{dy}{dt}\right)\)

So, \(\displaystyle \overrightarrow{v}=(2bt, 2b)\)

Therefore, the velocity when \(\displaystyle t=7\) is \(\displaystyle \overrightarrow{v}=(14b, 2b)\)
 
Last edited:
Dec 2014
31
0
Australia
You sub t into the y once you have made t the subject in terms of x, you don't have to as far as i know find t in terms of y