# Find an equation of the tangent line at the given point on the curve

Given the parametric equations, find an equation of the tangent line at the given point on the curve.

## Homework Statement

Find an equation of the tangent line at each given point on the curve:

x = 2cotΘ and y=2sin$^{2}θ$ at point ($\frac{-2}{\sqrt{3}}$,$\frac{3}{2})$

## Homework Equations

dy/dx = dy/dθ/dx/dθ

## The Attempt at a Solution

$\frac{-2}{\sqrt{3}} = 2cotθ$ and $\frac{3}{2}=2sin^{2}θ$

$\frac{-1}{\sqrt{3}}=\frac{cosθ}{sinθ }$ and $\frac{\sqrt{3}}{2}=sinθ$

letting θ = $\frac{2\pi}{3}$ for each function yields the point ($\frac{-2}{\sqrt{3}}$,$\frac{3}{2})$

dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $(\frac{2sin^{2}θ}{2cotθ})'$

= $-2sin^{3}θcosθ$

at θ= $\frac{2\pi}{3}$

slope of tangent = 9/8

so m = (y-y)/(x-x)

9/8 = $\frac{(y-\frac{3}{2}}{x-(-\frac{-2}{\sqrt{3}}}$

y = $\frac{9}{8} + 15/4$

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
PeroK
Homework Helper
Gold Member
2020 Award
dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $(\frac{2sin^{2}θ}{2cotθ})'$

= $-2sin^{3}θcosθ$
Try converting this derivative back to a function of x and y.

Last edited:
1 person
I don't think this can be converted into a function in terms of x and y. Why would I have to use the formula dy/dy = dy/dθ/dx/dθ if a function could just be found in terms of x and y from the start?

To create one function in terms of x and y, I would have to solve one of the parametric equations for θ and then substitute that result into the other parametric equation. That's the only way I can fathom coming up with a function in terms of x and y.

PeroK
Homework Helper
Gold Member
2020 Award
You don't think that dy/dx looks a bit like $xy^2$?

Last edited:
1 person
PeroK
Homework Helper
Gold Member
2020 Award
I don't think this can be converted into a function in terms of x and y. Why would I have to use the formula dy/dy = dy/dθ/dx/dθ if a function could just be found in terms of x and y from the start?
$x^2 = \frac{4cos^2(\theta)}{sin^2(\theta)} = \frac{8(1-sin^2(\theta))}{y}$

$= \frac{8-4y}{y} = \frac{8}{y} - 4$

$\frac{8}{y} = x^2 + 4 \ ∴ \ y = \frac{8}{x^2 + 4}$

So, you could have done it this way!

1 person
Chestermiller
Mentor
Given the parametric equations, find an equation of the tangent line at the given point on the curve.

## Homework Statement

Find an equation of the tangent line at each given point on the curve:

x = 2cotΘ and y=2sin$^{2}θ$ at point ($\frac{-2}{\sqrt{3}}$,$\frac{3}{2})$

## Homework Equations

dy/dx = dy/dθ/dx/dθ

## The Attempt at a Solution

$\frac{-2}{\sqrt{3}} = 2cotθ$ and $\frac{3}{2}=2sin^{2}θ$

$\frac{-1}{\sqrt{3}}=\frac{cosθ}{sinθ }$ and $\frac{\sqrt{3}}{2}=sinθ$

letting θ = $\frac{2\pi}{3}$ for each function yields the point ($\frac{-2}{\sqrt{3}}$,$\frac{3}{2})$

dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $(\frac{2sin^{2}θ}{2cotθ})'$

= $-2sin^{3}θcosθ$

at θ= $\frac{2\pi}{3}$

slope of tangent = 9/8

so m = (y-y)/(x-x)

9/8 = $\frac{(y-\frac{3}{2}}{x-(-\frac{-2}{\sqrt{3}}}$

y = $\frac{9}{8} + 15/4$

You did not calculate the derivative dy/dx correctly. This is where you made your mistake:
dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $(\frac{2sin^{2}θ}{2cotθ})'$
dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $\frac{(2sin^{2}θ)'}{(2cotθ)'}$

1 person
PeroK
Homework Helper
Gold Member
2020 Award
You did not calculate the derivative dy/dx correctly. This is where you made your mistake:
dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $(\frac{2sin^{2}θ}{2cotθ})'$
dy/dx = $\frac{dy/dθ}{dx/dθ}$ = $\frac{(2sin^{2}θ)'}{(2cotθ)'}$
Despite the notation and the appearance of doing the former, the OP did actually do the latter and differentiate the numerator and denominator separately - and correctly!

1 person
Chestermiller
Mentor
Despite the notation and the appearance of doing the former, the OP did actually do the latter and differentiate the numerator and denominator separately - and correctly!
Oh yes. I see you're right.

Chet

1 person
NascentOxygen
Staff Emeritus
and $\frac{\sqrt{3}}{2}=sinθ$
......
I haven't scrutinized this closely, apart from verifying you had differentiated the quotient correctly, as has one other poster.

You say answer doesn't match, do you mean your answer isn't the same as the answer given in the book? But might there be more than one "answer"? For example, there is often an angle in another quadrant which satisfies cos and sin. And when you took the square root to find sin, you haven't allowed for the possibility it's the negative of (√3)/2?

Once you have what you think are candidates for the answer, you need to check that each does satisfy the original equations.

1 person
You don't think that dy/dx looks a bit like $xy^2$?
PeroK, no, unfortunately I don't think dy/dx looks any bit like xy^2.

I think that $\frac{d}{dx}\frac{1}{3}xy^{3}$ looks a bit like xy$^{2}$

I don't see what you are implying by comparing dy/dx to xy^2.

Hi Lebombo!

The calculated slope is incorrect. :)

1 person
I haven't scrutinized this closely, apart from verifying you had differentiated the quotient correctly, as has one other poster.

You say answer doesn't match, do you mean your answer isn't the same as the answer given in the book? But might there be more than one "answer"? For example, there is often an angle in another quadrant which satisfies cos and sin. And when you took the square root to find sin, you haven't allowed for the possibility it's the negative of (√3)/2?

Once you have what you think are candidates for the answer, you need to check that each does satisfy the original equations.
The answer for this problem in the back of the book is $3\sqrt{3}x - 8y + 18 = 0$

I can try the angle from another quadrant, the only other possibility might be 10pi/6 in the 4th quadrant. I'll try it and see what I come up with.

Last edited:
PeroK
Homework Helper
Gold Member
2020 Award
PeroK, no, unfortunately I don't think dy/dx looks any bit like xy^2.

I think that $\frac{d}{dx}\frac{1}{3}xy^{3}$ looks a bit like xy$^{2}$

I don't see what you are implying by comparing dy/dx to xy^2.
We have:

$\frac{dy}{dx} = -2sin^3(θ)cos(θ) = -2sin^4(θ)\frac{cos(θ)}{sin(θ)} = -2sin^4(θ)cot(θ)$

$= -2(\frac{y^2}{4})(\frac{x}{2}) = -\frac{xy^2}{4}$

Now you can just plug in the values for x and y without worrying about θ.

You have to be a bit more open-minded about how to solve these problems.

1 person
I can try the angle from another quadrant, the only other possibility might be 10pi/6 in the 4th quadrant. I'll try it and see what I come up with.
You don't really need to look for any other angle. Your work is fine except when you substitute the values of ##\sin\theta## and ##\cos\theta## in ##-2\sin^3\theta \cos\theta##. What is ##(-\sqrt{3}/2)^3##?

1 person
Pranav-Arora,

Thanks, when I saw your first point, I immediately went back to double check the evaluation for the slope.

I noted that I did not calculate $\frac{\sqrt{3^{3}}}{2^{3}}$ correctly.

I was going to respond back to you with the correct equation in hand, but you beat me to the punch. I am in the process of forming what I hope will be he right answer, but so far, one term is still not matching up.

So far, I have $3\sqrt{3}x - 8y + 34/3 = 0$

According to the book, the 34/3 term should actually be 18.

Last edited:
So far, I have $3\sqrt{3}x - 8y + 34/3 = 0$

According to the book, the 34/3 term should actually be 18.
Can you show your steps? I do reach the answer given by the book.

1 person
These are the steps I've made:

$\frac{\sqrt{27}}{8}=\frac{y-2/3}{x+2/\sqrt{3}}$

$\frac{\sqrt{27}}{8}(x+\frac{2}{\sqrt{3}})= y- 2/3$

$\sqrt{27}x + 2\frac{\sqrt{27}}{\sqrt{3}}+\frac{16}{3}=8y$

$3\sqrt{3}x + 18/3 + 16/3 -8y = 0$

$3\sqrt{3}x + 34/3 - 8y = 0$

Last edited:
These are the steps I've made:

$\frac{\sqrt{27}}{8}=\frac{y-2/3}{x+2/\sqrt{3}}$

$\frac{\sqrt{27}}{8}(x+\frac{2}{\sqrt{3}}= y- 2/3$

$\sqrt{27}x + 2\frac{\sqrt{27}}{\sqrt{3}}+\frac{16}{3}=8y$

$3\sqrt{3}x + 18/3 + 16/3 -8y = 0$

$3\sqrt{3}x + 34/3 - 8y = 0$
The ordinate of given point is 3/2, not 2/3. :)

1 person
Fantastic, thanks so much.

Fantastic, thanks so much.