Name: Increasing Functions
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: The GDP of India is either increasing or decreasing, depending on a variety of factors. Let’s understand what an increasing function is, by using derivatives...

Question

Prove that

y = a x^{3}, x^{2} + 3 y^{2} = b^{2}

are orthogonal.

Accepted Answer

Given curves are $y = a x^{3}, x^{2} + 3 y^{2} = b^{2}$ .

If two curves are orthogonal to each other, then the product of their slopes of tangents must be equal to $- 1$ .

Now, $y = a x^{3} . . . . . . (i)$

Differentiating it with respect to $x$ ,

$\frac{d y}{d x} = \frac{d}{d x} (a x^{3}) \Rightarrow {(\frac{d y}{d x})}_{1} = 3 a x^{2} . . . . . . . (ii)$

Now, $x^{2} + 3 y^{2} = b^{2}$

Differentiating it with respect to $x$ ,

$\Rightarrow \frac{d}{d x} (x^{2} + 3 y^{2}) = \frac{d}{d x} (b^{2}) \Rightarrow 2 x + 6 y (\frac{d y}{d x}) = 0 \Rightarrow 6 y (\frac{d y}{d x}) = - 2 x \Rightarrow (\frac{d y}{d x}) = - \frac{2 x}{6 y} = - \frac{x}{3 y} \Rightarrow {(\frac{d y}{d x})}_{2} = - \frac{x}{3 y} . . . . . . . (iii)$

From equations $(ii)$ and $(iii)$ ,

$\Rightarrow {(\frac{d y}{d x})}_{1} \times {(\frac{d y}{d x})}_{2} = 3 a x^{2} \times (- \frac{x}{3 y})$

$\Rightarrow {(\frac{d y}{d x})}_{1} \times {(\frac{d y}{d x})}_{2} = (- \frac{a x^{3}}{y})$

$\{∵ y = a x^{3}\}$

$\Rightarrow {(\frac{d y}{d x})}_{1} \times {(\frac{d y}{d x})}_{2} = (- \frac{y}{y})$

$\Rightarrow {(\frac{d y}{d x})}_{1} \times {(\frac{d y}{d x})}_{2} = - 1$

Hence, proved that given curves are orthogonal.

Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: Gujarat Board-2018

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Application of Derivatives, Exercise 1: Gujarat Board-2018

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