Finding the slope of a tangent line to polar curve

Question

Question: Find the slope of the tangent line to the graph of $r = e^\theta - 4$ at $\theta = \frac{\pi}{4}$.

$$x = r\cos \theta = (e^\theta - 4)\cos\theta$$

$$y = r\sin \theta = (e^\theta - 4)\sin\theta$$

$$\frac{dx}{d\theta} = -e^\theta\sin\theta + e^\theta\cos\theta + 4\sin\theta$$ $$\frac{dy}{d\theta} = e^\theta\cos\theta + e^\theta\sin\theta - 4\cos\theta$$

$$\frac{dy}{dx} = \frac{e^\theta(\cos\theta + \sin\theta) - 4\cos\theta}{e^\theta(\cos\theta - \sin\theta) + 4\sin\theta}$$

$$\frac{dy}{dx} = \frac{\sqrt{2}(e^{\frac{\pi}{4}}-2)}{2\sqrt{2}} = \frac{1}{2}e^{\frac{\pi}{4}} - 1$$

When I plugged this problem into Wolfram Alpha (http://www.wolframalpha.com/input/?i=slope+of+the+tangent+line+to+r+%3D+e%5E(theta)-4+at+theta%3D(pi%2F4)), it said that the answer was just $e^{\frac{\pi}{4}}$, so I'm confused where I went wrong in my steps. I tried looking over the arithmetic a couple of times but couldn't find an incorrect step.

Any pointers or help would be appreciated - thank you very much!

Answer 1

Your answer is correct. Wolframalpa only finding derivative of $e^{\theta}-4$ with respect to $\theta$ so answer is only $e^{\frac{\pi}{4}}$.

You can check with similar problems.

Answer 2

Your answer is definitely correct, it is a mistake on wolfram alpha's end. If you ask it to find $\frac{dy}{dx}$ having given it $y(\theta)$ and $x(\theta)$ - then it arrives at the same answer as you: http://www.wolframalpha.com/input/?i=find+dy%2Fdx+if+y+%3D+(e%5Etheta+-4)sintheta,+x%3D+(e%5Etheta+-+4)costheta