Week 5 Worksheet

  1. Find the derivatives of the following functions:

    1. \(f(x) = x^3 - 11x^2 + 7x + 9\).

      Solution

      \(f'(x) = 3x^2 - 22x + 7\).

    2. \(g(x) = -100\sqrt{x} - 11x^{2/3}\).

      Solution

      Rewrite \(g\) as \(g(x) = -100x^{1/2} - 11x^{2/3}\). Then \(g'(x) = -50x^{-1/2} - \frac{22}{3}x^{-1/3}\).

    3. \(h(x) = \frac{8}{x^5} - \frac{8}{x^4} + \frac{6}{x} + \sqrt{7}\).

      Solution

      Rewrite \(h\) as \(h(x) = 8x^{-5} - 8x^{-4} + 6x^{-1} + \sqrt{7}\). Then \(h'(x) = -40x^{-6} + 32x^{-5} - 6x^{-2}\).

    4. \(k(x) = (8x^2 - 4x)^2\).

      Solution

      Expand \(k\) as \(k(x) = 64x^4 - 64x^3 + 16x^2\). Then \(k'(x) = 256x^3 - 192x^2 + 32x\).

      We also could have used the product rule or chain rule and arrived at the same answer.

  2. On the way to section, a GSI walks east down University Ave. While walking, he moves at about a constant rate, but stops 3 times at intersections. After getting to campus, he continues to walk without stopping, but slows down on the way up the hill to Evans. He waits for packets to print, and then turns around and sprints west towards Barker.

    Let \(f(t)\) represent the GSI’s distance from his starting location. Sketch possible graphs of both \(f\) and \(f'\).

    Solution

    The graph of \(f\) should alternate between positive, constant slope and zero slope three times (walking and stopping at intersections). After that, it should be increasing but concave down (walking but slowing down on the hill). Then it should be zero slope again (waiting for papers). Last, it should be decreasing, and at a steeper slope than any of the previous parts (sprinting back). In any case, the graph of \(f\) should not have any jumps, since teleportation is impossible.

    Correspondingly, \(f'\) should alternate between a positive constant and zero, with either a jump or a very steep line connecting them (a jump would correspond to stopping/starting instantly, but in reality it takes some time to speed up and slow down). It should then start up at a positive constant and decrease to some smaller positive constant (corresponding to slowing down on the hill), then jump to zero (waiting for papers) and last, jump to some negative constant (sprinting back).

  3. It turns out that a company’s cost and revenue for producting \(x\) items is modeled by

    \[ C(x) = \frac{2}{x} \qquad R(x) = 2x - \frac{x^2}{5000} \]

    1. Find functions for marginal cost and marginal revenue.

      Solution

      Marginal cost and marginal revenue are given by \(C'(x) = -\frac{2}{x^2}\) and \(R'(x) = 2 - \frac{x}{2500}\).

    2. Use your answer to the previous part to find a function for marginal profit.

      Solution

      Marginal profit is thus \[ P'(x) = R'(x) - C'(x) = 2 - \frac{x}{2500} - \frac{2}{x^2} \]

    3. For what value of \(x\) is marginal profit 0? When is marginal profit positive? When is it negative? (You may use a calculator for this, or otherwise reason about approximate values.)

      Solution

      Using a caclulator/computer, we find that this has a single root at \(x \approx 5000\). Even without a calculator, we can verify that this makes sense, since as \(x\) gets very large, \(\frac{2}{x^2} \approx 0\), and \(5000\) is a solution of \(2 - \frac{x}{2500} = 0\). When \(x\) is larger than this root, \(P'(x)\) is negative, and when \(x\) is smaller, \(P'(x)\) is positive.

    4. How many products should the company produce?

      Solution

      This means that profit is increasing when \(x < 5000\) and decreasing when \(x > 5000\), so it attains a maximum when \(x = 5000\). Thus the company should make and sell \(5000\) items to maximize profits.

  4. Suppose the position of an object moving along a straight line is given by \(s(t) = t^3 - 2t + 1\). What is the velocity when \(t = 2\)? What is the acceleration when \(t = 2\)?

    Solution

    Velocity is \(s'(t) = 3t^2 - 2\), so \(s'(2) = 3(2)^2 - 2(2) = 8\). Acceleration is \(s''(t) = 6t\), so \(s''(2) = 12\).

  5. The typical velocity \(v\) (in centimeters per second) of a marine organism of length \(l\) (in centimeters) is given by \(v = 2.69 \cdot l^{1.86}\). Find the rate of change of velocity with respect to length at \(l = 1.25\). What are the units of this quantity?

    Solution

    The rate of change is \(\frac{dv}{dl} = 2.69 \cdot 1.86 l^{0.86}\). At \(l = 1.25\), this gives \(\frac{dv}{dl} \approx 6.0619\).

    The units of \(\frac{dv}{dl}\) are the units of \(v\) (centimeters per second) divided by the units of \(l\) (centimeters), or

    \[ \frac{\text{centimeters/seconds}}{\text{centimeters}} = \frac{1}{\text{seconds}} \]