Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

A variable

Δ ABC

in the

x y

plane has its orthocentre at vertex

B,

a fixed vertex '

A

' at the origin and the third vertex '

C

' restricted to lie on the parabola

y = 1 + \frac{7 x^{2}}{36} .

The point

B

starts at the point

(0, 1)

at time

t = 0

and moves upward along the

y

- axis at a constant velocity of

2 cm / \sec

. How fast is the area of the triangle increasing when

t = \frac{7}{2} \sec .

Accepted Answer

Let $x y$ -coordinate of point $C$ is $h$ .

Given that point $C$ lies on the parabola $y = 1 + \frac{7 x^{2}}{16}$ ,

so $B$ -coordinate of point $(0, 1)$ will be $1 + \frac{7 h^{2}}{16}$ .

Hence $2 cm / \sec$ .

Also given that at $t = 0$ , $B$ starts at the point $(0, 1)$ and moves upward along the $y$ -axis at a constant velocity $2 cm / \sec$ .

Let co-ordinate of point $B$ is $B (0, k)$ .

$\frac{d y}{d t} = 2 cm / \sec$ or $\frac{d k}{d t} = 2 cm / \sec$

&

From the above figure $x y$

$\Rightarrow \frac{d k}{d t} = \frac{7 (h)}{18} \frac{d h}{d t}$

$\Rightarrow \frac{d h}{d t} = (\frac{36}{7 h}) \dots (i)$

Area of triangle,

$A = \frac{1}{2} \times Base \times Height$

$\Rightarrow A = \frac{1}{2} \times (h) \times (1 + \frac{7 h^{2}}{36}) \dots (ii)$

As point $B$ moves upwards along the $y$ -axis at a constant velocity of $2 cm / \sec$ .

So after $\frac{7}{2} \sec$ ., change in $y$ -coordinate $= 7$

Hence after $\frac{7}{2} \sec$ ., $B (0, 8)$ at $t = \frac{7}{2} \sec$ .

$\Rightarrow 1 + \frac{7 h^{2}}{36} = 8$ ; $k = 1 + \frac{7 h^{2}}{36}$

$\Rightarrow h^{2} = 36$

$\Rightarrow h = 6$

So, $C (6, 8)$ at $t = \frac{7}{2} \sec$ .

Now using equation $(ii)$

$\frac{d A}{d t} = \frac{1}{2} \{h (\frac{14}{36} \cdot h) + (1 + \frac{7 h^{2}}{36})\} (\frac{d h}{d t})$

$\Rightarrow \frac{d A}{d t} = \frac{1}{2} \{36 (\frac{14}{36}) + (8)\} (\frac{36}{7 h})$ ; $\frac{d h}{d t} = \frac{36}{7 h}$

$\Rightarrow \frac{d A}{d t} = \frac{1}{2} (14 + 8) (\frac{6}{7}) = \frac{66}{7}$

Important Questions on Application of Derivatives

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