Parametric equations help

Original post by ShadowFax895

(a) At what values of t does the curve have vertical or horizontal tangent? (not quite sure what to do here)

differentiate

What is the gradient for a horizontal line, and for a vertical line

Reply 2

OP

Original post by TenOfThem

differentiate

What is the gradient for a horizontal line, and for a vertical line

when we differentiate we use the dy/dx = dy/dt * dt/dx formula right?

i know that the gradient of a horizontal line is 0, but i thought the gradient of a vertical line was undefined, what is the gradient of a vertical line?

(edited 10 years ago)

Reply 3

brianeverit

9

Original post by ShadowFax895

hi

can someone help me with parametric equations, i have no idea how to do part (a) in this question from my textbook

Consider the curve given by the parametric equations
x = 2αcost − αcos2t and y = 2αsint−αsin2t, 0 ≤ t < 2π
.

(a) At what values of t does the curve have vertical or horizontal tangent? (not quite sure what to do here)

(b) Find the tangent line of the curve above when t = π/4.
(assuming you derive the equation and plug in the t value, then find x and y and make a linear equation out of it right)

(c) Sketch the curve for a = 1
(i think i know how to sketch it, just sub in alpha=1 and plot x on y right)

Horizontal tangent if dy/dt=0, vertical if dx/dt=0

Reply 4

Original post by ShadowFax895

when we differentiate we use the dy/dx = dy/dt * dt/dx formula right?

i know that the gradient of a horizontal line is 0, but i thought the gradient of a vertical line was undefined, what is the gradient of a vertical line?

It is undefined

Your gradient function will be a fraction

You will want the denominator to =0

Reply 5

OP

Original post by TenOfThem

It is undefined

Your gradient function will be a fraction

You will want the denominator to =0[/QUOT

Im still pretty confused,
first we use the dy/dx = dy/dt * dt/dx formula to differentiate it
we find the horizontal value by dy/dx = 0,
and the vertical value by..............

also when we differentiate do we use the product rule also, and the t value's will be in terms of alpha?

Reply 6

Original post by ShadowFax895

Im still pretty confused,
first we use the dy/dx = dy/dt * dt/dx formula to differentiate it
we find the horizontal value by dy/dx = 0,
and the vertical value by..............

also when we differentiate do we use the product rule also, and the t value's will be in terms of alpha?

For a vertical line the gradient needs to be

\dfrac{something}{0}

So you need the denominator of the gradient to be 0

Reply 7

OP

just to make sure im not making any stupid mistakes, can someone please tell me the derivative of x = 2αcost − αcos2t
the a is confusing me, how do i diferentiate with a and t, is it implicit differentiation, substitution or what?

(edited 10 years ago)

Reply 8

davros

16

Original post by ShadowFax895

just to make sure im not making any stupid mistakes, can someone please tell me the derivative of x = 2αcost − αcos2t
the a is confusing me, how do i diferentiate with a and t, is it implicit differentiation, substitution or what?

Surely the 'a' is just a constant so it's just standard differentiation?

Reply 9

Kvothe the Arcane

20

Original post by ShadowFax895

just to make sure im not making any stupid mistakes, can someone please tell me the derivative of x = 2αcost − αcos2t
the a is confusing me, how do i diferentiate with a and t, is it implicit differentiation, substitution or what?

\dfrac{d}{dx}a\times f(x)=a \times f'(x)

Provided a represent a constant and isn't a variable then it won't affect the differentiation :biggrin:

(edited 10 years ago)

Reply 10

OP

No idea what the last post was about,

so if x = 2αcost − αcos2t and y = 2αsint−αsin2t,

dy/dt = 2acos(t) +2cos(2t)

dx/dt = -2asin(t) + 2asin(2)
(could someone check if its right)

then what do we do from there?

we can put dy/dt over dx/dt to get dy/dx, then what?

Reply 11