第 12 类 RD Sharma 解决方案 - 第 11 章微分 - 练习 11.7 |设置 3

问题 21. 如果$x=a\left(\frac{1+t^2}{1-t^2}\right)$和$y=\frac{2t}{1-t^2}$ ， 找$\frac{dy}{dx}$

解决方案：

Here,

$x=a\left(\frac{1+t^2}{1-t^2}\right)$

Differentiate it with respect to t using chain rule,

$\frac{dx}{dt}=a\left[\frac{(1+t^2)\frac{d}{dt}(1+t^2)-(1+t^2)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(2t)-(1+t^2)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{2t-2t^2+2t+2t^3}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{4at}{(1-t^2)^2}\ \ \ \ \ ......(1)$

And,

$y=\frac{2t}{1-t^2}$

Differentiate it with respect to t using quotient rule,

$\frac{dy}{dt}=a\left[\frac{(1-t^2)\frac{d}{dt}(t)-(t)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(1)-(t)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{1-t^2+2t}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{2(1+t^2)}{(1-t^2)}\ \ \ \ \ ......(2)$

问题 22. 查找$\frac{dy}{dx}$ , 如果 y = 12(1 – cos t), x = 10(t – sin t), $-\frac{\pi}{2}<t<\frac{\pi}{2}$

解决方案：

It is given that, 

y = 12(1 – cos t),

x = 10(t – sin t)

Therefore,

$\frac{dx}{dt}=\frac{d}{dt}[10(t-sin\ t)]\\ =10.\frac{d}{dt}(t-sin\ t)\\ =10(1- cos\ t)$

$\frac{dy}{dt}=\frac{d}{dt}[12(t-cos\ t)]\\ =12.\frac{d}{dt}(1-cos\ t)\\ =12(0- (-sin\ t)\\ =12sin\ t$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12sin\ t}{10(1-cos\ t)}\\ =\frac{12\times2sin\frac{t}{2}\times cos\frac{t}{2}}{10\times2\ sin^2\frac{t}{2}}\\ =\frac{6}{5}\ cot\ \frac{t}{2}$

问题 23. 如果 x = a( θ – sin θ ) 且 y = a(1 – cos θ)，求$\frac{dy}{dx}$ ,在 θ = $\frac{\pi}{3}$

解决方案：

Here,

x = a(θ – sin θ)

and

y = a(1 – cos θ)

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}[a(θ-sin\ θ]\\ =a(1-cos\ θ)$

$\frac{dy}{dθ}=\frac{d}{dθ}[a(1+cos\ θ]\\ =a(-sin\ θ)$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-asin\ θ}{a(1-cos\ θ)}|_{θ=\frac{\pi}{3}}\\ =-\frac{sin\frac{\pi}{3}}{1-cos\frac{\pi}{3}}\\ =\frac{\frac{\sqrt3}{2}}{1-\frac{1}{2}}=-\sqrt3$

问题 24. 如果 x = a sin 2t (1 + cos 2t) 并且 y = b cos 2t (1 – cos 2t)，证明在 t = $\frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}$

解决方案：

Consider the given functions,

x = a sin 2t (1 + cos 2t)

and 

y = b cos 2t (1 – cos 2t)

Write again the functions,

x = a sin 2t + $\frac{a}{2}$sin 4t

Differentiate the above function with respect to t,

$\frac{dx}{dt}=2a\ cos\ 2t+2a\ cos\ 4t\ \ \ \ \ ....(1)\\$

y = b cos 2t (1 – cos 2t)

y = b cos 2t – b cos2 2t

$\frac{dy}{dx}=-2b\ sin\ 2t+2b\ cos\ 2t\ sin\ 2t\\=-2b\ sin\ 2t+b\ sin\ 4t\ \ \ \ ....(2)$

From equation (1) and (2)

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2b\ sin\ 2t+b\ sin\ 4t}{2a\ cos\ 2t+2a\ cos\ 4t}\\ \therefore\frac{dy}{dx}|_{\frac{\pi}{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}|_{t=\frac{\pi}{4}}=\frac{-2b}{-2a}=\frac{b}{a}$

问题 25. 如果 x = cos t (3 – 2cos ² t) 和 y = sin t (3 – 2 sin ² t)，求$\frac{dy}{dx}$在 t = $\frac{\pi}{4}$

解决方案：

Here, the given function:

x = cos t (3 – 2cos2t)

x = cos t – 2cos3t

$\frac{dx}{dt}=-3sin\ t+6cos^2t\ \ \ \ ......(1)$

y = sin t (3 – 2 sin2t)

y = 3cos t – 2sin3t

$\frac{dy}{dt}=3cos\ t-6sin^2tcos\ t\ \ \ \ .....(2)\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3cos\ t-6sin^tcos\ t}{-3sin\ t+6cos^2t\ sin\ t}\\ =\frac{3cos\ t(1-2sin^2t)}{3sin\ t(2cos^2t-1)}\\ =cot\ t\frac{(1-2(1-cos^2t))}{(2cos^2t-1)}\\ =cot\ t\\ \frac{dy}{dx}|_{\frac{\pi}{4}}=cot\frac{\pi}{4}=1$

问题 26. 如果$x=\frac{1+log\ t}{t^2}$ , $y=\frac{3+2log\ t}{t}$找$\frac{dy}{dx}$

解决方案：

Here,

$x=\frac{1+log\ t}{t^2}$

 and

$y=\frac{3+2log\ t}{t}$

$\frac{dx}{dt}=\frac{t^2\left(\frac{1}{t}\right)-(1+log\ t)(2t)}{t^4}\\ =\frac{t-2t-2tlog\ t}{t^4}\\ =\frac{-2log\ t-1}{t^3}$

$\frac{dy}{dt}=\frac{t\left(\frac{2}{t}\right)-(-3+2log\ t)(1)}{t^2}\\ =\frac{2-3-2tlog\ t}{t^2}\\ =\frac{-2log\ t-1}{t^2}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-2log\ t-1}{t^2}}{\frac{-2log\ t-1}{t^3}}=t$

问题 27. 如果 x = 3sin t – sin3t, y = 3cos t – cos3t，求$\frac{dy}{dx}\ at\ t=\frac{\pi}{3}$

解决方案：

x = 3sin t – sin3t

and,

y = 3cos t – cos3t

$\frac{dx}{dt}=3cos\ t-3cos3t\\ \frac{dy}{dt}=-3sin\ t+3sin3t\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3sin\ t+3sin3t}{3cos\ t-3cos3t}$

When, $t=\frac{\pi}{3}$

$\frac{dy}{dx}=\frac{-3sin(\frac{\pi}{3})+3sin(\pi)}{3cos(\frac{\pi}{3})-3cos(\pi)}=\frac{-3\times\frac{\sqrt3}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt3}$

问题 28. 如果$sin\ x=\frac{2t}{1+t^2}$ , $tan\ y=\frac{2t}{1-t^2}$找$\frac{dy}{dx}$

解决方案：

$sin\ x=\frac{2t}{1+t^2}$

and,

$tan\ y=\frac{2t}{1-t^2}$

$\Rightarrow x=sin^{-1}\left(\frac{2t}{1+t^2}\right)$

and 

$\Rightarrow y=tan^{-1}\left(\frac{2t}{1-t^2}\right)$

$\frac{dx}{dt}=\frac{1}{\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}}\times\frac{2(1+t^2)-(2t)(2t)}{(1+t^2)^2}\\ \frac{dx}{dt}=\frac{2}{1+t^2}\\ \frac{dy}{dt}=\frac{1}{\left(\frac{2t}{1-t^2}\right)2+1}\times\frac{2(1-t^2)-(2t)(-2t)}{(1-t^2)^2}\\ \frac{dy}{dt}=\frac{2}{1+t^2}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{(1+t^2)}}{\frac{2}{(1+t^2)}}=1$