# Help with solving parametric equation

• 53Mark53
In summary, the parametric curve x=5cos^7(t) and y=5sin^7(t) can be written in cartesian form as x^(2/7) + y^(2/7) = 1.

## Homework Statement

Consider the following parametric curve:

x=5cos^7(t)

y=5sin^7(t)

Write it in cartesian form, giving your answer as an equation of the form F(x,y)=c for some function F and some constant c.

## The Attempt at a Solution

[/B]
I know that sin^2(t)+cos^2(t) = 1 but I don't think this will be much help to figure this out and I am also unsure how I would find the constant

Any help would be much appreciated

53Mark53 said:
I know that sin^2(t)+cos^2(t) = 1 but I don't think this will be much help to figure this out
Why not? How are ##x## and ##y## related to ##\sin^2(t)## and ##\cos^2(t)##?

Fightfish said:
Why not? How are ##x## and ##y## related to ##\sin^2(t)## and ##\cos^2(t)##?

would that mean x^2+y^2 = 1

which would mean

(5cos^7(t))^2 +(5sin^7(t))^2 = 1

25cos^9(t) +25sin^9(t) = 1

But what would I do now?

53Mark53 said:
would that mean x^2+y^2 = 1
Nope, that is not correct. Try to express ##\sin^2(t)## in terms of ##y## and ##\cos^2(t)## in terms of ##x##.

Fightfish said:
Nope, that is not correct. Try to express ##\sin^2(t)## in terms of ##y## and ##\cos^2(t)## in terms of ##x##.
would that mean

y=1-cos^2(t)

x=1-sin^2(t)

53Mark53 said:
would that mean

y=1-cos^2(t)

x=1-sin^2(t)
No, why would you think that?
You already have ##x## and ##y## defined in your original post: ##x = 5 \cos^7(t)## and ##y = 5 \sin^7(t)##.
Can you manipulate ##x = 5 \cos^7(t)## to get ##\cos^2(t)## in terms of ##x##? And likewise for ##\sin^2(t)## in terms of ##y##.

53Mark53 said:

## Homework Statement

Consider the following parametric curve:

x=5cos^7(t)

y=5sin^7(t)

Write it in cartesian form, giving your answer as an equation of the form F(x,y)=c for some function F and some constant c.

## The Attempt at a Solution

[/B]
I know that sin^2(t)+cos^2(t) = 1 but I don't think this will be much help to figure this out and I am also unsure how I would find the constant

Any help would be much appreciated
To start: Solve
##x=5\cos^7(t) ##​
for ##\ \cos(t)\ .##

SammyS said:
To start: Solve
##x=5\cos^7(t) ##​
for ##\ \cos(t)\ .##

cos(t)=(x/5)^(7/2)

Is this correct?

53Mark53 said:
cos(t)=(x/5)^(7/2)

Is this correct?
No. The exponent is wrong.

SammyS said:
No. The exponent is wrong.

cos(t)=(x/5)^(1/7)

What would i do now?

53Mark53 said:
cos(t)=(x/5)^(1/7)

What would i do now?
That's better.

Do similar for y.

53Mark53
SammyS said:
That's better.

Do similar for y.