Calculus I

Enter in your name, I need to know who to give the grades to.

Let \(f(x) = \sqrt{x+1}, g(x)=x^2\), and \(h(x) = (f \circ g)(x) = f(g(x))\).

1. Define these function in Desmos, but hide the graphs of \(f\) and \(g\).
2. Let \(a \in [-5,5]\) and graph the point \((a,h(a))\).
3. Graph the tangent line to graph of \(h\) at \(x=a\).
4. Make a cell which calculates \(h'(a)\), and another which calculates \(f'(g(a))g'(a)\). Verify that these are the same values for various \(a\).

1. Graph \(x^2+y^2=1\), and the point \(P = (-3/5,4/5)\).
2. Find and graph the line tangent to the graph in part 1 at the point \(P\).

1. Graph \(y\cos(y + x + x^2) = x^3\) and the point \(Q = (0,\pi/2)\).
2. Find and graph the line tangent to the graph in part 1 at the point \(Q\).

1. In Desmos define \(j(\theta) = \cos(3\theta)\), but hide the graph. Then define \(r = j(\theta)\), for \(\theta \in [0,\pi]\). Show this graph.
2. Graph \(r = 1\). What do you notice about these two graphs? You are graphing in polar coordinates (make sure Desmos is in radian mode).
3. Find and graph the three points where the two graphs intersect.
4. For each of the points in part 3, graph a line tangent to the graph in part 1 at the said point. There are two ways to do this, see the corresponding hints below. (The conversion between rectangular and polar coordinates is \(x=r\cos \theta\) and \(y=r\sin\theta\)).
   • Hint 1: The three lines will also be tangent to the graph of the circle.

   • Hint 2: Notice that \(r = j(\theta)\), thus:

     \begin{align*} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} \\ &= \frac{j'(\theta)\sin\theta + j(\theta)\cos\theta} {j'(\theta)\cos\theta - j(\theta)\sin\theta}. \end{align*}