Question 6.27: Use the first and second derivatives to determine the curvat......
Use the first and second derivatives to determine the curvature of each of the following curves:
(a) y = 3x^4 + 20 (b) Q = \frac {25}{L}
in terms of whether:
(i) The curve is increasing or decreasing.
(ii) Curvature is concave up or concave down.
(iii) Curvature is convex or concave towards the origin (in first quadrant only).
Make a rough sketch of each curve.
In each case, find the first and second derivatives. Then check whether the derivatives are positive or negative.
(a) y = 3x^4 + 20
(i) y′= 12x³\begin{cases} \text {when} x > 0, y^{\prime} > 0 → y \text {is increasing as x increases} \\ \text {when} x < 0, y^{\prime} > 0 → y \text {is increasing as x increases} \end{cases}
(ii) y″ = 36x² positive → concave up for any value of x.
(iii) The curve is graphed in the first quadrant of Figure 6.32, and it may be described as convex towards the origin.
(iv) Plot about four or five points as shown in Table 6.15 to find the general shape of the curve.
(b) Q = \frac {25}{L} = 25L^{−1}
(i) Q′ = 25(−L^{−2}) = −25L^{−2} = – \frac {25}{L²}
→ negative for all L → curve is falling
(ii) Q″ = -25(−2L^{−3}) = 50L^{−3} = – \frac {50}{L³}
→ positive for all L > 0 → concave up
(iii) In the first quadrant this curve is described as convex towards the origin.
(iv) To plot the curve, calculate some points as shown in Table 6.16 which are then plotted in Figure 6.33.
Again note division by zero when x = 0, hence a vertical asymptote.
| Table 6.15 Points for Figure 6.32 | ||||||
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y = 3x^4 + 20 | 20 | 23 | 68 | 263 | 788 | 1895 |
| Table 6.16 Points for Figure 6.33 | ||||||
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| Q = \frac {25}{L} | ? | 25 | 12.5 | 8.33 | 6.25 | 5 |
