2.10.6 Differentiation Short Questions (Question 22 - 25)

Question 22:

Given that

y = \frac{3}{4} x^{2}

, find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:

\begin{array}{l} y = \frac{3}{4} x^{2} \\ \frac{d y}{d x} = (2) \frac{3}{4} x = \frac{3}{2} x \\ δ y = 47.7 - 48 = - 0.3 \\ Approximate change in x to y \\ \frac{δ x}{δ y} \approx \frac{d x}{d y} \\ δ x = \frac{d x}{d y} \times δ y \\ δ x = \frac{2}{3 x} \times (- 0.3) \\ δ x = \frac{2}{3 (8)} \times (- 0.3) \leftarrow \begin{array}{l} y = 48 \\ \frac{3}{4} x^{2} = 48 \\ x^{2} = 64 \\ x = 8 \end{array} \\ δ x = - 0.025 \end{array}

Question 23:

Given that

y = 15 x + \frac{24}{x^{3}}

(a) Find the value of

\frac{d y}{d x}

when x = 2,

(b) Express in terms of k, the approximate change in y when x changes from 2 to

2 + k, where k is a small change.

Solution:
(a)

\begin{array}{l} y = 15 x + \frac{24}{x^{3}} \\ y = 15 x + 24 x^{- 3} \\ \frac{d y}{d x} = 15 - 72 x^{- 4} \\ \frac{d y}{d x} = 15 - \frac{72}{x^{4}} \\ When x = 2 \\ \frac{d y}{d x} = 15 - \frac{72}{2^{4}} = 10.5 \end{array}

(b)

\begin{array}{l} Approximate change in y to x in terms of k, \\ \frac{δ y}{δ x} \approx \frac{d y}{d x} \\ δ y = \frac{d y}{d x} \times δ x \\ δ y = 10.5 \times (2 + k - 2) \\ δ y = 10.5 k \end{array}

Question 24:

If the radius of a circle increases from 4 cm to 4.01 cm, find the approximate increase in the area.

Solution:

\begin{array}{l} Area of circle, A = π r^{2} \\ \frac{d A}{d r} = 2 π r \\ Approximate increase in the area to radius, \\ \frac{δ A}{δ r} \approx \frac{d A}{d r} \\ δ A = \frac{d A}{d r} \times δ r \\ δ A = (2 π r) \times (4.01 - 4) \\ δ A = [2 π (4)] \times (0.01) \\ δ A = 0.08 π {cm}^{2} \end{array}

Question 25:

Given that y =3t+ 5t² and x = 5t –1.

(a) Find

\frac{d y}{d x}

in terms of x,

(b) If x increases from 5 to 5.01, find the small increase in t.

Solution:

\begin{array}{l} y = 3 t + 5 t^{2} \\ \frac{d y}{d t} = 3 + 10 t \\ x = 5 t - 1 \\ \frac{d x}{d t} = 5 \end{array}

(a)

\begin{array}{l} \frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} \\ \frac{d y}{d x} = (3 + 10 t) \times \frac{1}{5} \\ \frac{d y}{d x} = \frac{3 + 10 (\frac{x + 1}{5})}{5} \leftarrow \begin{array}{l} x = 5 t - 1 \\ t = \frac{x + 1}{5} \end{array} \\ \frac{d y}{d x} = \frac{3 + 2 x + 2}{5} \\ \frac{d y}{d x} = \frac{5 + 2 x}{5} \end{array}

(b)

\begin{array}{l} Small increase in t to x, \\ δ t = \frac{d t}{d x} \times δ x \\ δ t = \frac{1}{5} \times (5.01 - 5) \\ δ t = 0.002 \end{array}

2.10.6 Differentiation Short Questions (Question 22 – 25)