# Cartesian equation question

#### Giovanni55

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

#### DenisB

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)

#### piccolo95

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:

#### Giovanni55

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

*

Last edited: