Q Find the curve whose polar subtangent is constant

[himanshu121]

I have done this thing::
$$\frac{rsin(\theta)}{k}= - \frac{r+tan(\theta) \frac{dr}{d\theta}}{-rtan\theta + \frac{dr}{d\theta}}$$

Rearranging and solving i got

$$rsin \theta \frac{d(rcos\theta)}{d\theta}=-k\frac{d(rsin\theta)}{d\theta}$$

on further solving i got
$$rcos\theta= -kln(rsin\theta) +c$$

But the answer at the back is ::
:: The Reciprocal Spiral ::
:: r(\theta - \alpha) = c::

Where I went wrong ::Pls Help

 

[thepatientmental]

It seems like you have made a small mistake in your calculations. The correct solution for the curve with constant polar subtangent is indeed the reciprocal spiral, which can be written as r(\theta - \alpha) = c.

To get to this solution, you need to start with the equation you have rearranged and solved:

rsin\theta \frac{d(rcos\theta)}{d\theta}=-k\frac{d(rsin\theta)}{d\theta}

But instead of solving for rcos\theta, you need to solve for r, as the equation for the reciprocal spiral is in terms of r, not rcos\theta.

So, let's start with your equation and solve for r:

rsin\theta \frac{d(rcos\theta)}{d\theta}=-k\frac{d(rsin\theta)}{d\theta}

Dividing both sides by rsin\theta, we get:

\frac{d(rcos\theta)}{d\theta}=-k\frac{d(rsin\theta)}{d\theta} \cdot \frac{1}{rsin\theta}

Using the product rule for differentiation, we can rewrite the left side as:

\frac{d(rcos\theta)}{d\theta}=cos\theta \cdot \frac{dr}{d\theta} - r\sin\theta

And the right side can be rewritten as:

-k\frac{d(rsin\theta)}{d\theta} \cdot \frac{1}{rsin\theta} = -k \cdot \frac{d(sin\theta)}{d\theta} \cdot \frac{1}{sin\theta} - k \cdot \frac{d(r)}{d\theta} \cdot \frac{1}{r}

Combining these, we get:

cos\theta \cdot \frac{dr}{d\theta} - r\sin\theta = -k \cdot \frac{d(sin\theta)}{d\theta} \cdot \frac{1}{sin\theta} - k \cdot \frac{d(r)}{d\theta} \cdot \frac{1}{r}

Next, we can rearrange this equation to solve for dr/d\theta:

cos\theta \cdot \frac{dr}{d\theta} + k \cdot