1. ## Parametric solving

Hi again;

I have been given the following questions:

Code:
Find the equation of the (i) tangent and (ii) normal lines to the curve given parametrically
by x = t2 + t, y = t2 − t at the point corresponding to t = 0.
now this is what I have done so far:

dx/dt = 2t
dy/dt = 2t
THEREFORE
dy/dx = 2t/2t

But am not sure if I am correct, or if I am what do I do next?

Thanks

2. $\displaystyle \frac{dy}{dt}=2t-1$

$\displaystyle \frac{dx}{dt}=2t+1$

When t=0, dy/dx = -1

$\displaystyle \frac{dy}{dx}=m=\frac{2t-1}{2t+1}$

(x,y)=(0,0)

Now use $\displaystyle y-y_{1}=m(x-x_{1})$

$\displaystyle y-0 = -1(x-0)$

y=-x

Now, it's easy to find the normal.

3. Originally Posted by galactus

Now use $\displaystyle y-y_{1}=m(x-x_{1})$
Is that like y=mx+c?

ALSO

so the normal would have gradient -1/m
so the normal would be -1/-1
giving y = 1x+c?

4. Originally Posted by taurus
Is that like y=mx+c?

ALSO

so the normal would have gradient -1/m
so the normal would be -1/-1
giving y = 1x+c?
the formula he used is called the point-slope formula. it is very commonly used, you should note it

yes, the normal's gradient is the negative inverse of the gradient of the tangent line

5. for normal in my case i got m = 1
therefore y = 1x+c => y = x+c

How do i get c then? In the answers i have its just x?

6. Originally Posted by taurus
for normal in my case i got m = 1
therefore y = 1x+c => y = x+c

How do i get c then? In the answers i have its just x?
the line goes through the origin. you can use the point slope form as galalctus showed you, or you could just realize that if a line passes through the origin, the y-intercept is zero

7. Originally Posted by galactus
$\displaystyle \frac{dy}{dt}=2t-1$

$\displaystyle \frac{dx}{dt}=2t+1$

When t=0, dy/dx = -1

$\displaystyle \frac{dy}{dx}=m=\frac{2t-1}{2t+1}$

(x,y)=(0,0)

Now use $\displaystyle y-y_{1}=m(x-x_{1})$

$\displaystyle y-0 = -1(x-0)$

y=-x

Now, it's easy to find the normal.
How do you know what values of x and y to use in this point slope equation?