Parametric solving

**taurus** · Oct 23rd 2007, 02:02 PM

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

**galactus** · Oct 23rd 2007, 02:26 PM

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

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

When t=0, dy/dx = -1

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

(x,y)=(0,0)

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

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

y=-x

Now, it's easy to find the normal.

**taurus** · Oct 23rd 2007, 02:58 PM

Originally Posted by galactus

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

Is that like y=mx+c?
could you please explain that

ALSO

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

**Jhevon** · Oct 23rd 2007, 03:24 PM

Originally Posted by taurus

Is that like y=mx+c?
could you please explain that

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

**taurus** · Oct 23rd 2007, 03:51 PM

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?

**Jhevon** · Oct 23rd 2007, 03:54 PM

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

**taurus** · Oct 30th 2007, 08:25 AM

Originally Posted by galactus

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

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

When t=0, dy/dx = -1

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

(x,y)=(0,0)

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

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

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?

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