I need to find the equation for the line tangent to the curve:
$$ x = 2\cos t, y = 2\sin t, t = \frac \pi4 $$
I also need to find the value of: $$ \frac{d^2y}{dx^2} $$
I'm not too sure what to do.. I got the slope of -1 by doing $\frac{dy}{dx}$ which let to $-cott$ and plugging in $\frac \pi4$
Any help on getting the rest of the problem done would be greatly appreciated.
Disclaimer: My high school's Calculus I course does not teach parametric functions, and I haven't taken the course for even a full semester yet.
Finding the tangent line at $\frac {\pi}{4}$
One definition for cosine and sine is the coordinates of the unit-circle. A unit-circle is by definition said to be centered around the origin and to have a radius length of 1 unit. A slightly manipulated version of the standard formula for circles around the origin ($\sqrt{x^2+y^2}=r$) gives us: $$\sqrt{x^2 + y^2} = 1$$ If x and y ($\cos(x)$ and $\sin(x)$ respectively) are multiplied by 2 then: $$\sqrt{(2x)^2 + (2y)^2}$$ $$\sqrt{4x^2 + 4y^2}$$ $$\sqrt{4( x^2 + y^2 )}$$ $$2\sqrt{x^2+y^2}$$ Substituting $\sqrt{x^2+y^2}$ with r based upon the above equation gives us: $$2r$$ Which means that our new radius is 2 units long. The equation for a circle with a radius size of 2 units that is centered around the origin is: $$x^2 + y^2 = 4$$ Now we can take the derivative with respect to x: $$2x + 2y*y'(x) = 0$$ Solving for $y'(x)$ gives: $$y'(x) = -\frac xy$$ At a $45°$ angle, $\cos(x) = \sin(x)$, thus, $x = y$. Therefore $y'(\frac {\pi}{4}) = -1$. To find the tangent line, you now just need the y-intercept. This can be found by using the equation $y=mx+b$ like so: $$\sqrt{2}=-\sqrt{2} + b$$ $$b = 2\sqrt{2}$$ Now we put it all together, again using $y=mx+b$, to obtain the tangent line: $$y = -x + 2\sqrt{2}$$
For visual confirmation of the above tangent line, here's a graph from Wolfram Alpha:
Finding a function for the second derivative:
By the quotient rule of derivatives, $y''(x)$ is: $$\frac {y-y'(x)*x}{y^2}$$ Subsitute $y'(x)$ with $-\frac {x}{y}$: $$y''(x) = \frac {y + \frac {x^2}{y}}{y^2}$$ Simplify: $$y''(x) = \frac {x^2 + y^2}{y^3}$$ We can optionally simplify further by using our circle equation $x^2 + y^2 = 4$: $$y''(x) = \frac {4}{y^3}$$ Finally, upon substitution of $2\sin(t)$ for y, we get: $$y''(x) = 2\csc(x)^3$$