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Understanding Parametric equations

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
I have a few trivial questions regarding finding equations of the tangent line:

1) Find an equation of the tangent line to the parametric curve:

x = 2 sin 2t y= 2 sin t

at the point \(\displaystyle (\sqrt{3}, 1)\)

The textbook says that point "corresponds to the parameter value \(\displaystyle t= \frac{pi}{6}\)"

How do they know that? After testing for a few minutes, I noticed that tan-1\(\displaystyle (\sqrt{3}/1) \) = \(\displaystyle t= \frac{pi}{6}\) Is that how they determined it?

2) Similar question but
x = t sin t
y = t cos t
point (0, -pi)

Using \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), I got
\(\displaystyle \frac{(cos t -sint)}{(sin t + cos t)}\) = slope of tangent line


I then plugged the "0" from the point (0, -pi) into the equation...
(cos 0 - sin 0) / (sin 0 + cos 0) = 1

Is that right/am I on the right track?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What I would do for the first one is use the chain rule as follows:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\)

Now, using the given parametric equations, we find:

\(\displaystyle \frac{dy}{dx}=\frac{2\cos(t)}{4\cos(2t)}= \frac{\cos(t)}{2\cos(2t)}\)

To determine the value of $t$, we could use:

\(\displaystyle y(t)=2\sin(t)=1\,\therefore\,\sin(t)=\frac{1}{2}\)

\(\displaystyle x(t)=2\sin(2t)=\sqrt{3}\,\therefore\,\sin(2t)= \frac{\sqrt{3}}{2}\,\therefore\, \cos(t)=\frac{\sqrt{3}}{2}\)

With no given restriction on $t$, we could state that these conditions correspond to:

\(\displaystyle t=\frac{\pi}{6}+2k\pi=\frac{\pi}{6}(12k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

However, the value of $k$ will not affect the equation of the tangent line, so we may as well let $k=0$, and use:

\(\displaystyle t=\frac{\pi}{6}\)

So, now you may compute the slope of the tangent line, and then using the given point, apply these to the point-slope formula to get the equation of the tangent line.

What do you find?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...
2) Similar question but
x = t sin t
y = t cos t
point (0, -pi)

Using \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), I got
\(\displaystyle \frac{(cos t -sint)}{(sin t + cos t)}\) = slope of tangent line


I then plugged the "0" from the point (0, -pi) into the equation...
(cos 0 - sin 0) / (sin 0 + cos 0) = 1

Is that right/am I on the right track?
You are using an incorrect value for $t$...does $y(0)=-\pi$?
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
What I would do for the first one is use the chain rule as follows:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\)

Now, using the given parametric equations, we find:

\(\displaystyle \frac{dy}{dx}=\frac{4\cos(2t)}{2\cos(t)}= \frac{2\cos(2t)}{\cos(t)}\)

To determine the value of $t$, we could use:

\(\displaystyle y(t)=2\sin(t)=1\,\therefore\,\sin(t)=\frac{1}{2}\)

\(\displaystyle x(t)=2\sin(2t)=\sqrt{3}\,\therefore\,\sin(2t)= \frac{\sqrt{3}}{2}\,\therefore\, \cos(t)=\frac{\sqrt{3}}{2}\)

With no given restriction on $t$, we could state that these conditions correspond to:

\(\displaystyle t=\frac{\pi}{6}+2k\pi=\frac{\pi}{6}(12k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

However, the value of $k$ will not affect the equation of the tangent line, so we may as well let $k=0$, and use:

\(\displaystyle t=\frac{\pi}{6}\)

So, now you may compute the slope of the tangent line, and then using the given point, apply these to the point-slope formula to get the equation of the tangent line.

What do you find?
Is it not

\(\displaystyle \frac{dy}{dx}=\frac{2\cos(t)}{4\cos(2t)}= \frac{\cos(t)}{2\cos(2t)}\)

Since,

\(\displaystyle \frac{dy}{dt} = 2 cost \)
(dy/dt [2 sin t] = 2 cos t)

\(\displaystyle \frac{dx}{dt} = 4\cos(2t) \)
(dy/dt [2 sin 2t] = 2 cos(2t) * 2)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Is it not

\(\displaystyle \frac{dy}{dx}=\frac{2\cos(t)}{4\cos(2t)}= \frac{\cos(t)}{2\cos(2t)}\)...
Yes, I misread, and have fixed my post.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
You are using an incorrect value for $t$...does $y(0)=-\pi$?
No, I guess I will try to use the same logic as 1)

x(t) = t sin t

y(t) = t cos t

We want the point (0, -pi)

Hence,

x(t) = t sin t = 0
therefore, t = 0

y(t) = t cos t = -pi
????


Question 1)

Got stuck here:

\(\displaystyle x(t)=2\sin(2t)=\sqrt{3}\,\therefore\,\sin(2t)= \frac{\sqrt{3}}{2}\,\therefore\, \cos(t)=\frac{\sqrt{3}}{2}\)

How does sin 2t = cos t?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
There are other values of $t$ which make $\sin(t)=0$. As you go around the unit circle, what is the next one?
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
There are other values of $t$ which make $\sin(t)=0$. As you go around the unit circle, what is the next one?
That would be at pi or 180 degrees.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...
Question 1)

Got stuck here:

\(\displaystyle x(t)=2\sin(2t)=\sqrt{3}\,\therefore\,\sin(2t)= \frac{\sqrt{3}}{2}\,\therefore\, \cos(t)=\frac{\sqrt{3}}{2}\)

How does sin 2t = cos t?
\(\displaystyle \sin(2t)=2\sin(t)\cos(t)=\frac{\sqrt{3}}{2}\)

Now, from the other equation, we know \(\displaystyle 2\sin(t)=1\), so we may write:

\(\displaystyle \cos(t)=\frac{\sqrt{3}}{2}\)
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Correct! :D

And what is $y(\pi)$?
$y(\pi)$ = pi cos pi = -1 pi




- - - Updated - - -

\(\displaystyle \sin(2t)=2\sin(t)\cos(t)=\frac{\sqrt{3}}{2}\)

Now, from the other equation, we know \(\displaystyle 2\sin(t)=1\), so we may write:

\(\displaystyle \cos(t)=\frac{\sqrt{3}}{2}\)
Very neat, you used the first equation to aid with the second. Is it always like this, (that we use the first equation to aid with the second)?
 
Last edited:

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
What I would do for the first one is use the chain rule as follows:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\)

Now, using the given parametric equations, we find:

\(\displaystyle \frac{dy}{dx}=\frac{2\cos(t)}{4\cos(2t)}= \frac{\cos(t)}{2\cos(2t)}\)

To determine the value of $t$, we could use:

\(\displaystyle y(t)=2\sin(t)=1\,\therefore\,\sin(t)=\frac{1}{2}\)

\(\displaystyle x(t)=2\sin(2t)=\sqrt{3}\,\therefore\,\sin(2t)= \frac{\sqrt{3}}{2}\,\therefore\, \cos(t)=\frac{\sqrt{3}}{2}\)

With no given restriction on $t$, we could state that these conditions correspond to:

\(\displaystyle t=\frac{\pi}{6}+2k\pi=\frac{\pi}{6}(12k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

However, the value of $k$ will not affect the equation of the tangent line, so we may as well let $k=0$, and use:

\(\displaystyle t=\frac{\pi}{6}\)

So, now you may compute the slope of the tangent line, and then using the given point, apply these to the point-slope formula to get the equation of the tangent line.

What do you find?
Back to question number 1!

Now that we determine t = pi/6, we can plug it back into the slope of the tangent line

\(\displaystyle \frac{\cos(t)}{2\cos(2t)} = \frac{\cos(pi/6)}{2\cos(pi /3 )} \)

Now that simplifies to root 3 / 2

Final Step:

Y2 - y1 = m (X2 - X1)
y2 - 1 = root 3 / 2 * (x2 - root 3)
y = (root 3 / 2) x - (1/2)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$y(\pi)$ = pi cos pi = -1 pi

Finishing this question:

Plug this into the slope of the tangent line:

(cos pi - sin pi) / (sin pi + cos pi) = -1 / -1 = 1?

- - - Updated - - -



Very neat, you used the first equation to aid with the second. Is it always like this, (that we use the first equation to aid with the second)?

You need to find the value of $t$ that satisfies both equations with regards to the given point. For example, using the first equation:

\(\displaystyle t\sin(t)=0\)

We know then by the zero-factor property that:

\(\displaystyle t=k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

Now, using this value for $t$ in the second equation, we find:

\(\displaystyle k\pi\cos(k\pi)=-\pi\)

\(\displaystyle k\cos(k\pi)=-1\)

We now see that \(\displaystyle k=1\) is the only viable value of $k$.

As far as always being able to use one equation to aid in solving the other, I can't say this is the case, but we can use whatever we find in one to help with the other when possible because they are in essence simultaneous equations.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
I get it now! I guess we can use whatever tactic we want, as long as it satisfies both equations.


\(\displaystyle t=k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

What does Z refer to?

Where can I read more about the zero factor thing?

Also, I've edited the first post of this page with the correct answer :)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Back to question number 1!

Now that we determine t = pi/6, we can plug it back into the slope of the tangent line

\(\displaystyle \frac{\cos(t)}{2\cos(2t)} = \frac{\cos(pi/6)}{2\cos(pi /3 )} \)

Now that simplifies to root 3 / 2

Final Step:

Y2 - y1 = m (X2 - X1)
y2 - 1 = root 3 / 2 * (x2 - root 3)
y = (root 3 / 2) x - (1/2)
Yes, that looks correct.

I would refrain from mixing uppercase and lower case variables. I would state:

We have the slope: \(\displaystyle m=\frac{\sqrt{3}}{2}\)

and the point: \(\displaystyle (\sqrt{3},1)\)

hence application of the point-slope formula gives us the tangent line:

\(\displaystyle y-1=\frac{\sqrt{3}}{2}\left(x-\sqrt{3} \right)=\frac{\sqrt{3}}{2}x-\frac{3}{2}\)

Arranging this in slope-intercept form, we have:

\(\displaystyle y=\frac{\sqrt{3}}{2}x-\frac{1}{2}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I get it now! I guess we can use whatever tactic we want, as long as it satisfies both equations.


\(\displaystyle t=k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

What does Z refer to?

Where can I read more about the zero factor thing?

\(\displaystyle k\in\mathbb{Z}\) just means that $k$ is an arbitrary integer.

The zero-factor property just means that if your have:

\(\displaystyle ab=0\)

then you know the equation is true when either $a=0$ or $b=0$, you are probably familiar with it when solving a quadratic by factoring, where you equate each factor in turn to zero to find the roots.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Answer to question 1.


Plug this into the slope of the tangent line:


(cos pi - sin pi) / (sin pi + cos pi) = -1 / -1 = 1

putting it in point-slope formula gives us y = x - pi

y2 - y1 = m (x2 - x1
y + pi = m (x)
y = x - pi

Answer in the back of the textbook : y = (1/pi) x - pi

Did I make a mistake somewhere?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Answer to question 1.


Plug this into the slope of the tangent line:


(cos pi - sin pi) / (sin pi + cos pi) = -1 / -1 = 1

putting it in point-slope formula gives us y = x - pi

Answer in the back of the textbook : y = (1/pi) x - pi

Did I make a mistake somewhere?
Yes, check your differentiation again, recalling you need to apply the product rule.
 

Rido12

Well-known member
MHB Math Helper
Jul 18, 2013
715
Whoops, did it right on paper, but when I tried to type it on here, I mistakened "sint (t)" to be sin (t). Thanks a lot! You were of great help!