第 12 类 RD Sharma 解决方案 - 第 11 章微分 - 练习 11.7 |设置 1

问题 1. 查找$\frac{dy}{dx}$ , 当： x = 在²和 y = 2at

解决方案：

Given that x = at², y = 2at

So,

$\frac{dx}{dt}=\frac{d}{dt}(at^2)=2at\\ \frac{dy}{dt}=\frac{d}{dt}(2at)=2a$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}$

问题 2. 查找$\frac{dy}{dx}$ , 当： x = a(θ + sinθ) 和 y = a(1 – cosθ)

解决方案：

Here,

x = a(θ + sinθ)

Differentiating it with respect to θ,

$\frac{dx}{dθ}=a(1+cosθ)\ \ \ ..........(1)$

and,

y = a(1 – cosθ)

Differentiate it with respect to θ,

$\frac{dy}{dθ}=a(θ+sinθ)\\ \frac{dy}{dθ}=asinθ\ \ \ \ .....(2)$

Using equation (1) and (2),

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}\\ =\frac{asinθ}{a(1-cosθ)}\\ =\frac{\frac{2sinθ}{2}\frac{cosθ}{2}}{\frac{2sin^2θ}{2}},\ \ \ \ \ \ \ \left\{Since,\ 1-cosθ=\frac{2sin2θ}{2},\frac{2sinθ}{2}\frac{cosθ}{2}=sinθ\right\}\\ =\frac{dy}{dx}=\frac{tanθ}{2}$

问题 3. 查找$\frac{dy}{dx}$ , 当： x = acosθ 和 y = bsinθ

解决方案：

Then x = acosθ and y = bsinθ

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}(acosθ)=-asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(bsinθ)=bcosθ$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{bcosθ}{-asinθ}=-\frac{b}{a}cotθ$

问题 4. 查找$\frac{dy}{dx}$ , 当: x = a e ^Θ (sinθ -cosθ), y = ae ^Θ (sinθ +cosθ)

解决方案：

Here,

x = ae^Θ (sinθ – cosθ)

Differentiating it with respect to θ,

$\frac{dx}{dθ}=a[e^θ\frac{d}{dθ}(sinθ-cosθ)+(sinθ-cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ+sinθ)+(sinθ-cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θsinθ]\ \ \ \ \ \ .......(1)$

And,

y = ae^Θ(sinθ+cosθ)

Differentiating it with respect to θ

$\frac{dy}{dθ}=a[e^θ\frac{d}{dθ}(sinθ+cosθ)+(sinθ+cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ-sinθ)+(sinθ+cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θcosθ]\ \ \ \ \ \ .......(2)$

Dividing equation (2) by equation (1)

$\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(2e^θ cosθ)}{a(2e^θ sinθ)}\\ \frac{dy}{dx}=cotθ$

问题 5. 查找$\frac{dy}{dx}$ , 当： x = bsin ² θ 和 y = acos ² θ

解决方案：

Here,

x = bsin²θ and y = acos²θ

Then,

$\frac{d}{dθ}=\frac{d}{dθ}(bsin^2θ)=2bsinθ cosθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(acos^2θ)=-2acosθ sinθ \\ \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-2acosθ sinθ}{2bsinθ cosθ}=-\frac{a}{b}\\$

问题 6. 查找$\frac{dy}{dx}$ , 当： x = a(1 – cos θ) 和 y = a(θ +sinθ) 在 θ = $\frac{\pi}{2}$

解决方案：

Here,

x = a(1 – cosθ) and y = a(θ + sinθ)

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}[a(1-cosθ)]=asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}[a(θ +sinθ)]=a(1+cosθ)$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(1+cosθ)}{a(sinθ)}|_{θ =\frac{x}{2}}\\ =\frac{a(1+0)}{a}=1$

问题 7. 查找$\frac{dy}{dx}$ ， 什么时候： $x=\frac{e^t+e^{-t}}{2}$和$y=\frac{e^t-e^{-t}}{2}$

解决方案：

Here,

$x=\frac{e^t+e^{-t}}{2}$

Differentiate it with respect to t,

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

and,

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

Differentiating it with respect to t,

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

Dividing equation (2) and (1)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x}{y}\\ \frac{dy}{dt}=\frac{x}{y}$

问题 8. 查找$\frac{dy}{dx}$ ， 什么时候： $x=\frac{3at}{1+t^2}$和$y=\frac{3at^2}{1+t^2}$

解决方案：

Here,

$x=\frac{3at}{1+t^2}$

Differentiating it with respect to t using quotient rule,

$\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at)-3at\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(3a)-3at(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{3a+3at^2-6at^2}{(1+t^2)^2}\right]\\ =\left[\frac{3a-3at^2}{(1-t^2)^2}\right]\\ \frac{dx}{dt}=\frac{3a(1-t^2)}{(1+t^2)^2}\ \ \ \ ....(1)$

and,

$y=\frac{3at^2}{1+t^2}$

Differentiating it with respect to t using quotient rule,

$\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at^2)-3at^2\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(6at)-(3at^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{6at+6at^3-6at^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{6at}{(1+t^2)^2}\ \ \ \ ....(2)$

Dividing equation (2) by (1)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6at}{(1+t^2)^2}\times\frac{(1+t^2)^2}{3a(1-t^2)}\\ \frac{dy}{dt}=\frac{2t}{1-t^2}$

问题 9. 如果 x 和 y 通过方程参数连接，不消去参数，求$\frac{dy}{dx}$当： x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

解决方案：

The given equations are

x = a(cosθ +θ sinθ) and y = a(sinθ -θcosθ)

Then,

$\frac{dx}{dθ}=a\left[\frac{d}{dθ}cosθ +\frac{d}{dθ}(θ sinθ)\right]\\ =a\left[-sinθ +θ \frac{d}{dθ}(sinθ)+sinθ \frac{d}{dθ}(θ)\right]$

= a[-sinθ + θcosθ + sinθ] = aθcosθ

$\frac{dy}{dθ}=a\left[\frac{d}{dθ}sinθ +\frac{d}{dθ}(θ cosθ)\right]\\ =a\left[cosθ -\{θ \frac{d}{dθ}(cosθ)+cosθ \frac{d}{dθ}(θ)\}\right]$

= a[cosθ +θsinθ -cosθ]

= aθsinθ

Therefore,

$\frac{dy}{dx}=\frac{\left(\frac{dy}{dθ}\right)}{\left(\frac{dx}{dθ}\right)}=\frac{aθsinθ}{aθ cosθ}=tanθ$

问题 10. 查找$\frac{dy}{dx}$ ， 什么时候： $x=e^θ \left(θ +\frac{1}{θ}\right)$和$y=e^{-θ} \left(θ -\frac{1}{θ}\right)$

解决方案：

Here,

$x=e^θ \left(θ +\frac{1}{θ}\right)$

Differentiating it with respect to θ using product rule,

$\frac{dx}{dθ}=e^θ \frac{d}{dθ}\left(θ +\frac{1}{θ}\right)+\left(θ +\frac{1}{θ}\right)\frac{d}{dθ}(e^θ)\\ =e^θ \left(1-\frac{1}{θ^2}\right)+\left(\frac{θ ^2+1}{θ}\right)e^θ \\ =e^θ \left(\frac{θ ^2-1+θ ^3+θ}{θ ^2}\right)\\ \frac{dx}{dθ}=\frac{e^θ (θ ^3+θ ^2+θ -1)}{θ ^2}\ \ \ .....(1)$

and,

$y=e^θ \left(θ -\frac{1}{θ}\right)$

Differentiating it with respect to θ using product rule and chain rule

$\frac{dy}{dθ}=e^{-θ} \frac{d}{dθ}\left(θ -\frac{1}{θ}\right)+\left(θ -\frac{1}{θ}\right)\frac{d}{dθ}(e^{-θ})\\ =e^{-θ} \left(1+\frac{1}{θ^2}\right)+\left(θ -\frac{1}{θ}\right)e^{-θ} \\ =e^{-θ} \left(1+\frac{1}{θ ^2}-θ +\frac{1}{θ}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{θ ^2+1 -θ^3 +θ}{θ ^2}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{-θ ^3+θ ^2+θ +1}{θ ^2}\right)\ \ \ \ ......(2)$