lets-say-we-have-a-function-f-x-The-tangent-line-at-x-n-makes-an-angle-with-the-x-axis-What-is-the-rate-of-change-of-the-angle-as-we-change-the-value-of-x-Suppose-f-x-x-2-

FacebookTweetPin

Question Number 3707 by Filup last updated on 19/Dec/15

$$\mathrm{lets}\mathrm{say}\mathrm{we}\mathrm{have}\mathrm{a}\mathrm{function}f\left(x\right).$$$$\mathrm{The}\mathrm{tangent}\mathrm{line}\mathrm{at}x=n\mathrm{makes}\mathrm{an}$$$$\mathrm{angle}\theta \mathrm{with}\mathrm{the}x-\mathrm{axis}.$$$$$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{the}\mathrm{angle}$$$$\mathrm{as}\mathrm{we}\mathrm{change}\mathrm{the}\mathrm{value}\mathrm{of}x?$$$$$$$$\mathrm{Suppose}f\left(x\right)=x^2$$

Commented by Rasheed Soomro last updated on 19/Dec/15

$$tan\theta =f{}^{\prime }\left(x\right)=2x$$$$\theta ={\tan }^{-1}\left(2x\right)$$$$\frac{d\theta }{dx}=\frac{d}{dx}\left({\tan }^{-1}\left(2x\right)\right)$$$$=\frac{1}{1+{\left(2x\right)}^2}\times \frac{d}{dx}\left(2x\right)=\frac{2}{1+4x^2}$$$${\left(\frac{d\theta }{dx}\right)}_{x=n}=\frac{2}{1+4n^2}$$

Commented by prakash jain last updated on 19/Dec/15

$$\mathrm{Rate}\mathrm{of}\mathrm{change}\mathrm{of}\mathrm{angle}\left(\theta \right)\mathrm{wrt}x=\frac{d\theta }{dx}$$$$f\left(x\right)=x^2$$$$\tan \theta =2x$$$${\sec }^2\theta \frac{d\theta }{dx}=2$$$$(1+4x^2)\frac{d\theta }{dx}=2$$$$\frac{\mathrm{d}\theta }{\mathrm{d}x}=\frac{2}{1+4x^2}$$

Commented by 123456 last updated on 19/Dec/15

$$\mathrm{so}$$$$\tan \theta =\frac{df}{dx}$$$$\theta ={\tan }^{-1}\frac{df}{dx}$$$$\frac{d\theta }{dx}=\frac{d}{dx}{\tan }^{-1}\frac{df}{dx}$$$$=\frac{\frac{d^{2}f}{{dx}^2}}{1+{\left(\frac{df}{dx}\right)}^2}???$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin