Vector Valued Function Using (or misusing) Trig Identity

[kmacinto]

Homework Statement

The context of the problem is that it's a vector valued function (VVF) problem where I'm supposed to sketch a curve generated by a VVF. To make the sketching easier I'm supposed to convert a VVF to a real valued function so that I can take advantage of the shape of a curve covered in previous sections of the text in real valued function format.

Homework Equations

Sketch the curve represented by the vector valued function r(θ) and give the orientation of the curve. r(θ)=3sec(θ)i+2tan(θ)j.

The Attempt at a Solution

The real valued function that r(θ) converts to is a hyperbola given by the equation (x^2)/2=(y^2)/2+1

The steps given in the book are:
1) from r(θ) set:
x=3sec(θ), y=2tan(θ)

2) (x^2)/2=(y^2)/2+1 which is the solution.

My steps attempted are:
from: x=3sec(θ), y=2tan(θ)
1) divide x and y by the constants leaving:
x/3 = sec(θ) and y/2 = tan(θ)

2) equate both equations leaving:
x/3 + y/2 = sec(θ) + tan(θ)

3) square both sides leaving:
(x/3)^2 + (y/3)^2 = (sec(θ))^2 + (tan(θ))^2

and here's where I'm stuck...

Because both sides need a negative to work. For instance, for the (y/3)^2 to be on the right and positive it needs to be negative on the left and to get the right side to equal 1 using (tanθ)^2 + 1 = (secθ)^2, I need either the (tanθ)^2 or the (secθ)^2 to be negative where they're at now. Right?

Totally flustered!

Ken
PS. The attachment is the problem and given solution from the textbook.

 

[tiny-tim]

Hi Ken!

1) divide x and y by the constants leaving:
x/3 = sec(θ) and y/2 = tan(θ)

2) equate both equations leaving:
x/3 + y/2 = sec(θ) + tan(θ)

3) square both sides leaving:
(x/3)^2 + (y/3)^2 = (sec(θ))^2 + (tan(θ))^2

I don't understand what you're doing.

What is sec²θ - tan²θ ? ​

 

[HallsofIvy]

$(a+ b)^2\ne a^2+ b^2$