# Math Help - Find the length of tangent

1. ## Find the length of tangent

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

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

Then find the values of $x$ and $y$ where this tangent crosses the axes, by putting $y = 0$ and $x = 0$ 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 $\theta$, the length ...'. In which case, it's fairly straightforward. Find, in terms of $\theta$:

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

Then use Pythagoras Theorem to find the required length.

--------------------------- ---------------------------

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

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

$
= -tan \theta
$

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

there the slope of the tangent is 0

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

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

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

4. ## Length of tangent

Hello zorro
Originally Posted by zorro
--------------------------- ---------------------------

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

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

$
= -tan \theta
$

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

there the slope of the tangent is 0

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

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

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

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

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

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

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

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

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

Can you take it from there?

5. ## Problem encountered

Hello zorro

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

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

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

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

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

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

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

Can you take it from there?

After this i get this

$\Rightarrow x=a\cos\theta$

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

6. Hello zorro

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

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

7. ## Thanks mite

Hello zorro

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

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