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Q1:
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 2at^2, y = at^4$
If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 2at^2, y = at^4$
Solution :
Step 1: Identifying the Parametric System
We are given two equations where $x$ and $y$ are functions of a common parameter $t$:
- $x = 2at^2$
- $y = at^4$
Step 2: Differentiating $x$ with respect to $t$
Applying the Power Rule of differentiation ($\frac{d}{dt}[t^n] = nt^{n-1}$):
$\frac{dx}{dt} = \frac{d}{dt}(2at^2)$
$\frac{dx}{dt} = 2a \cdot \frac{d}{dt}(t^2)$
$\frac{dx}{dt} = 2a \cdot (2t)$
$\frac{dx}{dt} = 4at$
Step 3: Differentiating $y$ with respect to $t$
Applying the Power Rule of differentiation:
$\frac{dy}{dt} = \frac{d}{dt}(at^4)$
$\frac{dy}{dt} = a \cdot \frac{d}{dt}(t^4)$
$\frac{dy}{dt} = a \cdot (4t^3)$
$\frac{dy}{dt} = 4at^3$
Step 4: Calculating $\frac{dy}{dx}$
Using the relation $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{4at^3}{4at}$
Canceling the common terms $4a$ (assuming $a \neq 0$):
$\frac{dy}{dx} = \frac{t^3}{t}$
Applying the Quotient Rule for exponents ($\frac{t^m}{t^n} = t^{m-n}$):
$\frac{dy}{dx} = t^{3-1}$
$\frac{dy}{dx} = t^2$
Final Solution: $\frac{dy}{dx} = t^2$
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.6
- Q2: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a \cos \theta, y = b \cos \theta$
- Q3: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \sin t, y = \cos 2t$
- Q4: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = 4t, y = \frac{4}{t}$
- Q5: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta$
- Q6: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\theta - \sin \theta), y = a(1 + \cos \theta)$
- Q7: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$
- Q8: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a\left(\cos t + \log \tan\left(\frac{t}{2}\right)\right), y = a \sin t$
- Q9: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a \sec \theta, y = b \tan \theta$
- Q10: If $x$ and $y$ are connected parametrically by the equations, without eliminating the parameter, Find $\frac{dy}{dx}$. $x = a(\cos \theta + \theta \sin \theta), y = a(\sin \theta - \theta \cos \theta)$
- Q11: If $x = \sqrt{a^{\sin^{-1} t}}, y = \sqrt{a^{\cos^{-1} t}}$, show that $\frac{dy}{dx} = -\frac{y}{x}$
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.3
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.5
- Continuity and Differentiability Class 12 EXERCISE 5.7
Chapters in CBSE - Class 12 Mathematics
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