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Thread starter 14 years ago
#1
x = at^2 , y = at^3 (a constant)

I'm having problems here and wonder if it has something to do with the comment that a is constant.

t^2 = x/a

y= a [^sqrt(x/a)]^3

y = [sqrt(x/a)]/ a^2

ya^2 = sqrt(x/a)

and now I'm stuck

The book's answer is

ay^2 = x^3

and I don't see how to get there.

Thanks for your help

jorge
0
14 years ago
#2
x = at^2
y = at^3

cube the first, square the second:

x^3 = (a^3)(t^6)
y^2 = (a^2)(t^6)

rearrange both to isolate t:

t^6 = (x/a)^3
t^6 = (y/a)^2

equate to eliminate the t:

(x/a)^3 = (y/a)^2

as a is a constant, you can multiply through by a^3

(x^3) = ay^2
0
Thread starter 14 years ago
#3
(Original post by mik1w)
x = at^2
y = at^3

cube the first, square the second:

x^3 = (a^3)(t^6)
y^2 = (a^2)(t^6)

rearrange both to isolate t:

t^6 = (x/a)^3
t^6 = (y/a)^2

equate to eliminate the t:

(x/a)^3 = (y/a)^2

as a is a constant, you can multiply through by a^3

(x^3) = ay^2
Thank you very much

But the way I did it should have yielded the same result. No?

Jorge 0
14 years ago
#4
(Original post by Jorge)
But the way I did it should have yielded the same result. No?
Yeah, but you had a slight error.

y= a [sqrt(x/a)]^3
y = [sqrt(x/a)]/ a^2
Should be:
y= a [sqrt(x/a)]^3 = a[x^(3/2)/a^(3/2)] = x^(3/2)/a^(1/2)
=> a^(1/2)y = x^(3/2)
=> ay^2 = x^3 [by squaring both sides]
0
Thread starter 14 years ago
#5
Thanks

Can you help me with another one, please?

x = 3 - 2cosA , y= 5+2sinA

2cosA = 3-x , y-5= 2sinA

4cos^2A = (3-x)^2 , (y-5)^2 = 4 sin^2A

4(1-sin^2A) = (3-x)^2

4 -4sin^2A = (3-x)^2

4 sin^2A = - 1/4 (3-x)^2

sin^2A = [- 1/4 (3-x)^2]/4

so...

(y-5)^2 = 4 {[- 1/4 (3-x)^2]/4}

(y-5)^2 = [- 1/4 (3-x)^2]

and this is going wrong because the book's answer is

(x-3)^2 +(y-5)^2 = 4

What have I done wrong?

Thank you

jorge
0
14 years ago
#6
(Original post by Jorge)
Thanks

Can you help me with another one, please?

x = 3 - 2cosA , y= 5+2sinA

2cosA = 3-x , y-5= 2sinA

4cos^2A = (3-x)^2 , (y-5)^2 = 4 sin^2A

4(1-sin^2A) = (3-x)^2

4 -4sin^2A = (3-x)^2

4 sin^2A = - 1/4 (3-x)^2
How did you get to that?

4 - 4sin^2A = (3-x)^2

=> 4 - (3-x)^2 = 4sin^2A

Also, (y-5)^2 = 4 sin^2A

So (y-5)^2 = 4 - (3-x)^2

=> (y-5)^2 + (3-x)^2 = 4

Galois.
0
14 years ago
#7
4 -4sin^2A = (3-x)^2
4 sin^2A = - 1/4 (3-x)^2 .. Where'd the -1/4 come from? Should have been:
4sin^2A = 4 - (3-x)^2
4sin^2A + (3-x)^2 = 4
(y-5)^2 + (3-x)^2 = 4

Another method I prefer:
You have:
4cos^2A = (3-x)^2
4sin^2A = (y-5)^2
Add these 2 together:
4cos^2A + 4sin^2A = (3-x)^2 + (y-5)^2
4(cos^2A + sin^2A) = (3-x)^2 + (y-5)^2
4 = (3-x)^2 + (y-5)^2 [since sin^2A+cos^2A=1]
0
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