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

1. Jan 1, 2014

### Lebombo

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

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

dy/dx = dy/dθ/dx/dθ

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

Answer doesn't match. I must have made a mistake somewhere.

Last edited: Jan 1, 2014
2. Jan 1, 2014

### PeroK

Try converting this derivative back to a function of x and y.

Last edited: Jan 1, 2014
3. Jan 1, 2014

### Lebombo

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.

4. Jan 1, 2014

### PeroK

You don't think that dy/dx looks a bit like $xy^2$?

Last edited: Jan 1, 2014
5. Jan 1, 2014

### PeroK

$x^2 = \frac{4cos^2(\theta)}{sin^2(\theta)} = \frac{8(1-sin^2(\theta))}{y}$

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

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

So, you could have done it this way!

6. Jan 1, 2014

### Staff: Mentor

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

7. Jan 1, 2014

### PeroK

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!

8. Jan 1, 2014

### Staff: Mentor

Oh yes. I see you're right.

Chet

9. Jan 1, 2014

### Staff: Mentor

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.

10. Jan 2, 2014

### Lebombo

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.

11. Jan 2, 2014

### Saitama

Hi Lebombo!

The calculated slope is incorrect. :)

12. Jan 2, 2014

### Lebombo

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: Jan 2, 2014
13. Jan 2, 2014

### PeroK

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.

14. Jan 2, 2014

### Saitama

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 $(-\sqrt{3}/2)^3$?

15. Jan 2, 2014

### Lebombo

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: Jan 2, 2014
16. Jan 2, 2014

### Saitama

Can you show your steps? I do reach the answer given by the book.

17. Jan 2, 2014

### Lebombo

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: Jan 2, 2014
18. Jan 2, 2014

### Saitama

The ordinate of given point is 3/2, not 2/3. :)

19. Jan 2, 2014

### Lebombo

Fantastic, thanks so much.

20. Jan 2, 2014

### Saitama

Glad to help! :)