Short Questions (Question 17 – 19)

Posted on May 16, 2020 by Myhometuition

Question 17:

In the diagram above, the straight line PR is normal to the curve

$y = \frac{x^{2}}{2} + 1$ at Q. Find the value of k.

Solution:

$\begin{array}{l} y = \frac{x^{2}}{2} + 1 \\ \frac{d y}{d x} = x \\ At point Q, x -coordinate = 2, \\ Gradient of the curve, \frac{d y}{d x} = 2 \\ Hence, gradient of normal to the curve, P R = - \frac{1}{2} \\ \frac{3 - 0}{2 - k} = - \frac{1}{2} \\ 6 = - 2 + k \\ k = 8 \end{array}$

Question 18:
The normal to the curve y = x² + 3x at the point P is parallel to the straight line
y = –x + 12. Find the equation of the normal to the curve at the point P.

Solution:

Given normal to the curve at point P is parallel to the straight line y = –x + 12.

Hence, gradient of normal to the curve = –1.

As a result, gradient of tangent to the curve = 1

y = x² + 3x

$\frac{d y}{d x}$ = 2x + 3

2x + 3 = 1

2x = –2

x = –1

y = (–1)²+ 3(–1)

y = –2

Point P = (–1, –2).

Equation of the normal to the curve at point P is,

y – (–2) = –1 (x – (–1))

y + 2 = – x – 1

y = – x– 3

Question 19:

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:

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