# Thread: Find the length of tangent

1. ## Find the length of tangent

Find the length of the portion of the tangent to the curve $\displaystyle x = acos^3 \theta$, $\displaystyle y=a sin^3 \theta$ intercepted between coordinate axes.

2. ## Length of tangent

Hello zorro
Originally Posted by zorro
Find the length of the portion of the tangent to the curve $\displaystyle x = acos^3 \theta$, $\displaystyle y=a sin^3 \theta$ intercepted between coordinate axes.
I'll assume the question means 'Find, in terms of $\displaystyle \theta$, the length ...'. In which case, it's fairly straightforward. Find, in terms of $\displaystyle \theta$:

• $\displaystyle \frac{dx}{d\theta}$
• $\displaystyle \frac{dy}{d\theta}$
• Hence $\displaystyle \frac{dy}{dx}$
• The equation of the tangent at the point with parameter $\displaystyle \theta$

Then find the values of $\displaystyle x$ and $\displaystyle y$ where this tangent crosses the axes, by putting $\displaystyle y = 0$ and $\displaystyle x = 0$ in turn, in this equation.

Then use Pythagoras Theorem to find the required length.

3. ## is this right

Hello zorro

I'll assume the question means 'Find, in terms of $\displaystyle \theta$, the length ...'. In which case, it's fairly straightforward. Find, in terms of $\displaystyle \theta$:

• $\displaystyle \frac{dx}{d\theta}$
• $\displaystyle \frac{dy}{d\theta}$
• Hence $\displaystyle \frac{dy}{dx}$
• The equation of the tangent at the point with parameter $\displaystyle \theta$
Then find the values of $\displaystyle x$ and $\displaystyle y$ where this tangent crosses the axes, by putting $\displaystyle y = 0$ and $\displaystyle x = 0$ in turn, in this equation.

Then use Pythagoras Theorem to find the required length.

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i have got half of it can u tell me what to do next ..........

$\displaystyle \frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}$

$\displaystyle = -tan \theta$

f'(0) = - tan (0) = 0

there the slope of the tangent is 0

Now the eq of the tangent at pt $\displaystyle (x_0,y_0)$ is (0,0)

therefore eq is $\displaystyle (y - y_0) = m (x-x_0)$

ie (y-0)=0(x-0)
ie y=0

4. ## Length of tangent

Hello zorro
Originally Posted by zorro
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i have got half of it can u tell me what to do next ..........

$\displaystyle \frac{dx}{dy} = \frac{sin^2 \theta cos \theta}{- cos^2 \theta sin \theta}$

$\displaystyle = -tan \theta$

f'(0) = - tan (0) = 0

there the slope of the tangent is 0

Now the eq of the tangent at pt $\displaystyle (x_0,y_0)$ is (0,0)

therefore eq is $\displaystyle (y - y_0) = m (x-x_0)$

ie (y-0)=0(x-0)
ie y=0
Note that you have written $\displaystyle \frac{dx}{dy}$, instead of $\displaystyle \frac{dy}{dx}$. Also, perhaps you mis-read my post: I said to find the equation of the tangent at the point whose parameter is $\displaystyle \theta$, not $\displaystyle 0$.

So, using $\displaystyle \frac{dy}{dx} = -\tan \theta$, the equation of the tangent whose paramter is $\displaystyle \theta$ is:

$\displaystyle y- a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)$

Meets the $\displaystyle x$-axis where $\displaystyle y = 0$, and

$\displaystyle - a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)$

$\displaystyle \Rightarrow a \sin^2\theta\cos\theta=x-a\cos^3\theta$

$\displaystyle \Rightarrow x=a\cos\theta(\sin^2\theta+\cos^2\theta)=$ ?

Can you take it from there?

5. ## Problem encountered

Hello zorro

Note that you have written $\displaystyle \frac{dx}{dy}$, instead of $\displaystyle \frac{dy}{dx}$. Also, perhaps you mis-read my post: I said to find the equation of the tangent at the point whose parameter is $\displaystyle \theta$, not $\displaystyle 0$.

So, using $\displaystyle \frac{dy}{dx} = -\tan \theta$, the equation of the tangent whose paramter is $\displaystyle \theta$ is:

$\displaystyle y- a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)$

Meets the $\displaystyle x$-axis where $\displaystyle y = 0$, and

$\displaystyle - a\sin^3\theta = -\tan\theta(x-a\cos^3\theta)$

$\displaystyle \Rightarrow a \sin^2\theta\cos\theta=x-a\cos^3\theta$

$\displaystyle \Rightarrow x=a\cos\theta(\sin^2\theta+\cos^2\theta)=$ ?

Can you take it from there?

After this i get this

$\displaystyle \Rightarrow x=a\cos\theta$

But dont we have to find the length of the tangent .........Please advice

6. Hello zorro

We have found that the tangent meets the $\displaystyle x$-axis where $\displaystyle x = a\cos\theta$.

In a similar way, it meets the $\displaystyle y$-axis where $\displaystyle x = 0$, and therefore:
$\displaystyle y-a\sin^3\theta =-\tan\theta(0-a\cos^3\theta)$
$\displaystyle =a\sin\theta\cos^2\theta$
$\displaystyle \Rightarrow y=a\sin\theta(\sin^2\theta +\cos^2\theta)$
$\displaystyle =a\sin\theta$
So we want the distance between the points $\displaystyle (0,a\sin\theta)$ and $\displaystyle (a\cos\theta, 0)$; which is:
$\displaystyle \sqrt{a^2\cos^2\theta +a^2\sin^2\theta}$
$\displaystyle =a$

7. ## Thanks mite

Hello zorro

We have found that the tangent meets the $\displaystyle x$-axis where $\displaystyle x = a\cos\theta$.

In a similar way, it meets the $\displaystyle y$-axis where $\displaystyle x = 0$, and therefore:
$\displaystyle y-a\sin^3\theta =-\tan\theta(0-a\cos^3\theta)$
$\displaystyle =a\sin\theta\cos^2\theta$
$\displaystyle \Rightarrow y=a\sin\theta(\sin^2\theta +\cos^2\theta)$
$\displaystyle =a\sin\theta$
So we want the distance between the points $\displaystyle (0,a\sin\theta)$ and $\displaystyle (a\cos\theta, 0)$; which is:
$\displaystyle \sqrt{a^2\cos^2\theta +a^2\sin^2\theta}$
$\displaystyle =a$