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Parametric Equations >.< watch

1. Bah I never thought i'd be this hard!

2 questions which I can't answer with all my life:

1)The parametric equations of a curve

x= t + e^-t
y=1 - e^-t

Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

I found dy/dx but I can't make it into a log form >.<

2) Curve is defined parametrically

x= 2(1+costt)
y=4sin^2t

Find cartesian equation

At first I change 4sin^2t using the identities and made it into 4-4cos^2t
hence there are no sins and only cos.

BUT I CANT DO THE CARTESIAN BIT >.<

Help :'(
2. (Original post by shan735)
Bah I never thought i'd be this hard!

2 questions which I can't answer with all my life:

1)The parametric equations of a curve

x= t + e^-t
y=1 - e^-t

Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

I found dy/dx but I can't make it into a log form >.<

2) Curve is defined parametrically

x= 2(1+costt)
y=4sin^2t

Find cartesian equation

At first I change 4sin^2t using the identities and made it into 4-4cos^2t
hence there are no sins and only cos.

BUT I CANT DO THE CARTESIAN BIT >.<

Help :'(
For the first part, what do you get when you set dy/dx=1?

For the second one, note that when you want to get something in cartesian form, you want to eliminate all variables (namely 't' in this case) leaving only x and y terms. Rearrange your x-t equation for cos(t) and substitute this into your y-t equation.
3. (Original post by shan735)
Bah I never thought i'd be this hard!

2 questions which I can't answer with all my life:

1)The parametric equations of a curve

x= t + e^-t
y=1 - e^-t

Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

I found dy/dx but I can't make it into a log form >.<

2) Curve is defined parametrically

x= 2(1+costt)
y=4sin^2t

Find cartesian equation

At first I change 4sin^2t using the identities and made it into 4-4cos^2t
hence there are no sins and only cos.

BUT I CANT DO THE CARTESIAN BIT >.<

Help :'(

Q1.
dx/dt= 1-e^t
dy/dt=e^-t

so dy/dx = (e^-t)/(1-e^-t)

dy/dx=1
so

(e^-t)=(1-e^-t)

1=2e^-t
e^-t=1/2
takin logs
-t=ln 1/2
i.e
-t= -ln2 (as ln1/2=ln1-ln2)

so t=ln2

Q2.
x=2(1+cost)
y=4sin^2t

y=4(1-cos^2t)
x^2=4(1+2cost+cos^2t)

so y+x^2=8(1+cost)

i.e y+x^2=4x

so....

y=x(4-x)
4. (Original post by catjaum)
q1.
Dx/dt= 1-e^t
dy/dt=e^-t

so dy/dx = (e^-t)/(1-e^-t)

dy/dx=1
so

(e^-t)=(1-e^-t)

1=2e^-t
e^-t=1/2
takin logs
-t=ln 1/2
i.e
-t= -ln2 (as ln1/2=ln1-ln2)

so t=ln2

q2.
X=2(1+cost)
y=4sin^2t

y=4(1-cos^2t)
x^2=4(1+2cost+cos^2t)

so y+x^2=8(1+cost)

i.e y+x^2=4x

so....

Y=x(4-x)

edit:why neg dis???
5. (Original post by catjaum)
edit:why neg dis???
Probably because you didn't read the forum guidelines and posted a full solution straight off.

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Updated: March 22, 2011
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