find dy/dx

**action812** · July 27th 2010, 11:17 AM

consider the cardioid r=1+sin(t) remember that x=rcos(t) and y=rsin(t)

which means x = (1+sin(t))cos(t)
and y = (1+sin(t))sin(t)

now dy/dx : [ (dy/dt) // (dx/dt) ]

should i use chain rule on the x and y, or would i just distrube the cos and the sin? thanks

**Unknown008** · July 27th 2010, 11:26 AM

I think that it's easier to use the chain rule; less work, faster result.

**action812** · July 27th 2010, 11:46 AM

I got the dy/dx to be : (cos(t)*(2sin(t)+1)) / - (2sin(t) - 1)*(sin(t)+1) can anyone confirm this for me? TY

**Unknown008** · July 27th 2010, 10:35 PM

Hmm, okay...

I'm applying the chain rule.

$\frac{dx}{dt} = (1+sin(t))(-sin(t)) + cos(t).cos(t)$

$\frac{dy}{dt} = (1+sin(t))(cos(t)) + sin(t).cos(t)$

Then,

$\frac{dy}{dx} = \frac{(1+sin(t))(cos(t)) + sin(t).cos(t)}{(1+sin(t))(-sin(t)) + cos(t).cos(t)}$

$= \frac{cos(t)(1 + 2sin(t))}{-(sin(t) + 1)(2sin(t)-1)}$

Yes, it's correct.

**Soroban** · July 28th 2010, 04:30 AM

Hello, action812!

$\text}Consider the cardioid: }\;r\:=\:1+\sin\theta$

$\text{Find }\:\dfrac{dy}{dx}$

I would derive the formula first . . .

$\begin{array}{ccccccccc}x &=& r\cos\theta & \Rightarrow & \dfrac{dx}{d\theta} &=& \text{-}r\sin\theta + r'\cos\theta \\ \\[-4mm]<br /> y &=& r\sin\theta & \Rightarrow & \dfrac{dy}{d\theta} &=& r\cos\theta + r'\sin\theta \end{array}$

$\text{Hence: }\;\dfrac{dy}{dx} \;=\;\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta} } \;=\;\dfrac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}$ . [1]

We have: . $\begin{Bmatrix}r \:=\:1+\sin\theta \\ r' \:=\:\cos\theta \end{Bmatrix}$

Substitute into [1]:

. . $\dfrac{dy}{dx} \;=\;\dfrac{(1+\sin\theta)\cos\theta + \cos\theta\sin\theta}{\text{-}(1+\sin\theta)\sin\theta + \cos\theta\cos\theta}$

. . . . . $=\;\dfrac{\cos\theta + \sin\theta\cos\theta + \sin\theta\cos\theta}{\text{-}\sin\theta - \sin^2\!\theta + \cos^2\!\theta}$

. . . . . $=\;\dfrac{2\sin\theta\cos\theta + \cos\theta}{\cos^2\!\theta - \sin^2\!\theta - \sin\theta}$

. . . . . $=\;\dfrac{\sin2\theta + \cos\theta}{\cos2\theta - \sin\theta}$

**action812** · July 28th 2010, 09:43 AM

thanks guys!!

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