Exercise 5.6

---

Q.1 If

$$ x=2at^2,\qquad y=at^4, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=2at^2,\qquad y=at^4. $$

For parametric equations,

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}. $$

---

Step 1: Differentiate $x$ with respect to $t$.

$$ \begin{aligned} \frac{dx}{dt} &=\frac{d}{dt}(2at^2)\\ &=4at. \end{aligned} $$

---

Step 2: Differentiate $y$ with respect to $t$.

$$ \begin{aligned} \frac{dy}{dt} &=\frac{d}{dt}(at^4)\\ &=4at^3. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\[2mm] &= \frac{4at^3}{4at}\\[2mm] &=t^2. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=t^2.} $$

---

Q.2 If

$$ x=a\cos\theta,\qquad y=b\cos\theta, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=a\cos\theta,\qquad y=b\cos\theta. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}, $$

differentiate both equations with respect to $\theta$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{d\theta} &= -a\sin\theta. \end{aligned} $$

---

Step 2: Differentiate $y$.

$$ \begin{aligned} \frac{dy}{d\theta} &= -b\sin\theta. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-b\sin\theta}{-a\sin\theta}\\[2mm] &= \frac{b}{a}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=\frac{b}{a}.} $$

---

Q.3 If

$$ x=\sin t,\qquad y=\cos2t, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=\sin t,\qquad y=\cos2t. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, $$

differentiate both equations with respect to $t$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{dt} &= \cos t. \end{aligned} $$

---

Step 2: Differentiate $y$.

$$ \begin{aligned} \frac{dy}{dt} &= -2\sin2t. \end{aligned} $$

Using the identity

$$ \sin2t=2\sin t\cos t, $$

we get

$$ \begin{aligned} \frac{dy}{dt} &= -4\sin t\cos t. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-4\sin t\cos t}{\cos t}\\[2mm] &= -4\sin t. \end{aligned} $$

Since

$$ x=\sin t, $$

the derivative may also be written as

$$ \boxed{\frac{dy}{dx}=-4\sin t=-4x.} $$

Answer:

$$ \boxed{\frac{dy}{dx}=-4\sin t\;(\text{or }-4x).} $$

---

Q.4 If

$$ x=4t,\qquad y=\frac{4}{t}, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=4t,\qquad y=\frac{4}{t}=4t^{-1}. $$

For parametric equations,

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}. $$

---

Step 1: Differentiate $x$ with respect to $t$.

$$ \begin{aligned} \frac{dx}{dt} &=\frac{d}{dt}(4t)\\ &=4. \end{aligned} $$

---

Step 2: Differentiate $y$ with respect to $t$.

$$ \begin{aligned} \frac{dy}{dt} &=\frac{d}{dt}(4t^{-1})\\ &=-4t^{-2}\\ &=-\frac{4}{t^2}. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\[2mm] &= \frac{-\dfrac{4}{t^2}}{4}\\[2mm] &= -\frac{1}{t^2}. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=-\frac{1}{t^2}.} $$

---

Q.5 If

$$ x=\cos\theta-\cos2\theta,\qquad y=\sin\theta-\sin2\theta, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=\cos\theta-\cos2\theta, \qquad y=\sin\theta-\sin2\theta. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}, $$

differentiate both equations with respect to $\theta$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{d\theta} &= -\sin\theta + 2\sin2\theta. \end{aligned} $$

Using

$$ \sin2\theta=2\sin\theta\cos\theta, $$

we obtain

$$ \boxed{ \frac{dx}{d\theta} = \sin\theta(4\cos\theta-1). } $$

---

Step 2: Differentiate $y$.

$$ \begin{aligned} \frac{dy}{d\theta} &= \cos\theta - 2\cos2\theta. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \boxed{ \frac{dy}{dx} = \frac{\cos\theta-2\cos2\theta} {\sin\theta(4\cos\theta-1)}. } $$

---

Q.6 If

$$ x=a(\theta-\sin\theta), \qquad y=a(1+\cos\theta), $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=a(\theta-\sin\theta), \qquad y=a(1+\cos\theta). $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}, $$

differentiate both equations with respect to $\theta$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{d\theta} &= a(1-\cos\theta). \end{aligned} $$

---

Step 2: Differentiate $y$.

$$ \begin{aligned} \frac{dy}{d\theta} &= -a\sin\theta. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-a\sin\theta} {a(1-\cos\theta)}\\[2mm] &= -\frac{\sin\theta} {1-\cos\theta}. \end{aligned} $$

Using the identity

$$ \frac{\sin\theta}{1-\cos\theta} = \frac{1+\cos\theta}{\sin\theta}, $$

we obtain

$$ \boxed{ \frac{dy}{dx} = -\frac{1+\cos\theta}{\sin\theta} = -\cot\frac{\theta}{2}. } $$

Answer:

$$ \boxed{ \frac{dy}{dx} = -\frac{\sin\theta}{1-\cos\theta} = -\frac{1+\cos\theta}{\sin\theta} = -\cot\frac{\theta}{2}. } $$

---

Q.7 If

$$ x=\frac{\sin^3t}{\sqrt{\cos2t}}, \qquad y=\frac{\cos^3t}{\sqrt{\cos2t}}, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=\sin^3t(\cos2t)^{-\frac12}, \qquad y=\cos^3t(\cos2t)^{-\frac12}. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, $$

differentiate both equations with respect to $t$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{dt} &= 3\sin^2t\cos t(\cos2t)^{-\frac12} +\sin^3t\left(-\frac12\right) (\cos2t)^{-\frac32}(-2\sin2t)\\[2mm] &= \frac{3\sin^2t\cos t\cos2t+\sin^3t\sin2t} {(\cos2t)^{\frac32}}. \end{aligned} $$

Since

$$ \sin2t=2\sin t\cos t, $$

we get

$$ \begin{aligned} \frac{dx}{dt} &= \frac{\sin^2t\cos t \left(3\cos2t+2\sin^2t\right)} {(\cos2t)^{\frac32}}. \end{aligned} $$

Using

$$ \cos2t=\cos^2t-\sin^2t, $$

the bracket simplifies to

$$ 3\cos2t+2\sin^2t = 3\cos^2t-\sin^2t. $$

Hence,

$$ \boxed{ \frac{dx}{dt} = \frac{\sin^2t\cos t(3\cos^2t-\sin^2t)} {(\cos2t)^{3/2}}. } $$

---

Step 2: Differentiate $y$.

$$ \begin{aligned} \frac{dy}{dt} &= -3\cos^2t\sin t(\cos2t)^{-\frac12} +\cos^3t \left(-\frac12\right) (\cos2t)^{-\frac32} (-2\sin2t)\\[2mm] &= \frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin2t} {(\cos2t)^{\frac32}}. \end{aligned} $$

Using

$$ \sin2t=2\sin t\cos t, $$

we obtain

$$ \begin{aligned} \frac{dy}{dt} &= \frac{\sin t\cos^2t \left(-3\cos2t+2\cos^2t\right)} {(\cos2t)^{\frac32}}. \end{aligned} $$

Since

$$ -3\cos2t+2\cos^2t = -\cos^2t+3\sin^2t, $$

we get

$$ \boxed{ \frac{dy}{dt} = \frac{\sin t\cos^2t (3\sin^2t-\cos^2t)} {(\cos2t)^{3/2}}. } $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}}\\[2mm] &= \frac{ \sin t\cos^2t(3\sin^2t-\cos^2t) }{ \sin^2t\cos t(3\cos^2t-\sin^2t) }. \end{aligned} $$

Cancelling common factors,

$$ \boxed{ \frac{dy}{dx} = \frac{\cos t}{\sin t} \cdot \frac{3\sin^2t-\cos^2t} {3\cos^2t-\sin^2t}. } $$

---

Q.8 If

$$ x=a\left(\cos t+\log\tan\frac{t}{2}\right), \qquad y=a\sin t, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=a\left(\cos t+\log\tan\frac{t}{2}\right), \qquad y=a\sin t. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, $$

differentiate both equations with respect to $t$.

---

Step 1: Differentiate $x$.

Using the standard result,

$$ \frac{d}{dt} \left( \log\tan\frac{t}{2} \right) = \csc t, $$

we obtain

$$ \begin{aligned} \frac{dx}{dt} &= a(-\sin t+\csc t)\\[2mm] &= a\left( -\sin t+\frac1{\sin t} \right)\\[2mm] &= a\frac{1-\sin^2t}{\sin t}\\[2mm] &= a\frac{\cos^2t}{\sin t}. \end{aligned} $$

---

Step 2: Differentiate $y$.

$$ \frac{dy}{dt} = a\cos t. $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{a\cos t} {a\dfrac{\cos^2t}{\sin t}}\\[2mm] &= \frac{\sin t}{\cos t}\\[2mm] &= \tan t. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=\tan t.} $$

---

Q.9 If

$$ x=a\sec\theta, \qquad y=b\tan\theta, $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=a\sec\theta, \qquad y=b\tan\theta. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}, $$

differentiate both equations with respect to $\theta$.

---

Step 1: Differentiate $x$.

$$ \frac{dx}{d\theta} = a\sec\theta\tan\theta. $$

---

Step 2: Differentiate $y$.

$$ \frac{dy}{d\theta} = b\sec^2\theta. $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{b\sec^2\theta} {a\sec\theta\tan\theta}\\[2mm] &= \frac{b}{a} \cdot \frac{\sec\theta}{\tan\theta}\\[2mm] &= \frac{b}{a} \cdot \frac{1}{\sin\theta}. \end{aligned} $$

Using

$$ \sin\theta=\frac{\tan\theta}{\sec\theta}, $$

the derivative may also be written as

$$ \boxed{ \frac{dy}{dx} = \frac{b}{a}\csc\theta. } $$

---

Q.10 If

$$ x=a(\cos\theta+\theta\sin\theta), \qquad y=a(\sin\theta-\theta\cos\theta), $$

find $\dfrac{dy}{dx}$ without eliminating the parameter.

Solution:

Given,

$$ x=a(\cos\theta+\theta\sin\theta), \qquad y=a(\sin\theta-\theta\cos\theta). $$

For parametric equations,

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}} {\dfrac{dx}{d\theta}}. $$

---

Step 1: Differentiate $x$ with respect to $\theta$.

$$ \begin{aligned} \frac{dx}{d\theta} &= a\left( -\sin\theta + \frac{d}{d\theta}(\theta\sin\theta) \right). \end{aligned} $$

Using the product rule,

$$ \frac{d}{d\theta}(\theta\sin\theta) = \sin\theta+\theta\cos\theta. $$

Therefore,

$$ \begin{aligned} \frac{dx}{d\theta} &= a(-\sin\theta+\sin\theta+\theta\cos\theta)\\ &= a\theta\cos\theta. \end{aligned} $$

---

Step 2: Differentiate $y$ with respect to $\theta$.

$$ \begin{aligned} \frac{dy}{d\theta} &= a\left( \cos\theta - \frac{d}{d\theta}(\theta\cos\theta) \right). \end{aligned} $$

Again, using the product rule,

$$ \frac{d}{d\theta}(\theta\cos\theta) = \cos\theta-\theta\sin\theta. $$

Hence,

$$ \begin{aligned} \frac{dy}{d\theta} &= a(\cos\theta-\cos\theta+\theta\sin\theta)\\ &= a\theta\sin\theta. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{a\theta\sin\theta} {a\theta\cos\theta}\\[2mm] &= \tan\theta. \end{aligned} $$

Answer:

$$ \boxed{\frac{dy}{dx}=\tan\theta.} $$

---

Q.11 If

$$ x=\sqrt{a^{\sin^{-1}t}}, \qquad y=\sqrt{a^{\cos^{-1}t}}, $$

show that

$$ \frac{dy}{dx} = -\frac{y}{x}. $$

Solution:

Given,

$$ x=\sqrt{a^{\sin^{-1}t}} = a^{\frac12\sin^{-1}t}, $$ $$ y=\sqrt{a^{\cos^{-1}t}} = a^{\frac12\cos^{-1}t}. $$

Using

$$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}} {\dfrac{dx}{dt}}, $$

differentiate both equations with respect to $t$.

---

Step 1: Differentiate $x$.

$$ \begin{aligned} \frac{dx}{dt} &= a^{\frac12\sin^{-1}t} \cdot \log a \cdot \frac12 \cdot \frac1{\sqrt{1-t^2}}\\[2mm] &= \frac{x\log a} {2\sqrt{1-t^2}}. \end{aligned} $$

---

Step 2: Differentiate $y$.

Since

$$ \frac{d}{dt}(\cos^{-1}t) = -\frac1{\sqrt{1-t^2}}, $$

we obtain

$$ \begin{aligned} \frac{dy}{dt} &= a^{\frac12\cos^{-1}t} \cdot \log a \cdot \left( -\frac12 \cdot \frac1{\sqrt{1-t^2}} \right)\\[2mm] &= -\frac{y\log a} {2\sqrt{1-t^2}}. \end{aligned} $$

---

Step 3: Find $\dfrac{dy}{dx}$.

$$ \begin{aligned} \frac{dy}{dx} &= \frac{-\dfrac{y\log a}{2\sqrt{1-t^2}}} {\dfrac{x\log a}{2\sqrt{1-t^2}}}\\[2mm] &= -\frac{y}{x}. \end{aligned} $$

Hence proved,

$$ \boxed{\frac{dy}{dx}=-\frac{y}{x}.} $$

---