OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (I)

Exercise 8(I)

Find \( \frac { dy }{ dx } \) :

Question 1. If x = ct, y = \( \frac { c }{ t } \).
Answer: We are given the parametric equations:
\( x = ct \) ...(1)
\( y = \frac { c }{ t } \) ...(2)

First, we differentiate equation (1) with respect to \( t \):
\( \frac { dx }{ dt } = \frac { d }{ dt } (ct) = c \)

Next, we differentiate equation (2) with respect to \( t \):
\( \frac { dy }{ dt } = \frac { d }{ dt } \left( \frac { c }{ t } \right) = c \cdot (-1)t^{-2} = -\frac { c }{ t^2 } \)

Now, we use the chain rule for parametric differentiation to find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { -\frac { c }{ t^2 } }{ c } \)
\( \implies \frac { dy }{ dx } = -\frac { 1 }{ t^2 } \)
In simple words: We find how both x and y change with respect to t separately. Then, we divide the change in y by the change in x to find how y changes with respect to x. This method is used when x and y are both defined using a third variable, like t.

🎯 Exam Tip: Remember to differentiate both x and y with respect to the parameter (t in this case) and then divide the derivative of y by the derivative of x. Pay attention to signs during differentiation.

Question 2. x = \( \frac{1}{1-t^2} \), y = \( 1 + t^2 \).
Answer: We are given the parametric equations:
\( x = \frac{1}{1-t^2} \) ...(1)
\( y = 1+t^2 \) ...(2)

First, differentiate equation (1) with respect to \( t \). We can write \( x = (1-t^2)^{-1} \):
\( \frac { dx }{ dt } = -1(1-t^2)^{-2} \cdot \frac { d }{ dt }(1-t^2) \)
\( \implies \frac { dx }{ dt } = -(1-t^2)^{-2} (-2t) \)
\( \implies \frac { dx }{ dt } = \frac { 2t }{ (1-t^2)^2 } \)

Next, differentiate equation (2) with respect to \( t \):
\( \frac { dy }{ dt } = \frac { d }{ dt }(1+t^2) = 2t \)

Now, use the chain rule for parametric differentiation to find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { 2t }{ \frac { 2t }{ (1-t^2)^2 } } \)
\( \implies \frac { dy }{ dx } = 2t \cdot \frac { (1-t^2)^2 }{ 2t } \)
\( \implies \frac { dy }{ dx } = (1-t^2)^2 \)
In simple words: First, find how fast x and y change with t. Then, divide the change in y by the change in x. This gives us the final rate of change of y with respect to x.

🎯 Exam Tip: When differentiating expressions like \( \frac{1}{f(t)} \), it's often easier to rewrite them as \( [f(t)]^{-1} \) and use the chain rule, rather than the quotient rule. Be careful with negative signs.

Question 3. x = \( a\left(\frac{1+t^2}{1-t^2}\right) \), y = \( b\left(\frac{2 t}{1-t^2}\right) \).
Answer: We are given the parametric equations:
\( x = a \left( \frac{1+t^2}{1-t^2} \right) \) ...(1)
\( y = b \left( \frac{2t}{1-t^2} \right) \) ...(2)

First, differentiate equation (1) with respect to \( t \) using the quotient rule:
\( \frac { dx }{ dt } = a \left[ \frac{ (1-t^2) \frac{d}{dt}(1+t^2) - (1+t^2) \frac{d}{dt}(1-t^2) }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dx }{ dt } = a \left[ \frac{ (1-t^2)(2t) - (1+t^2)(-2t) }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dx }{ dt } = a \left[ \frac{ 2t - 2t^3 + 2t + 2t^3 }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dx }{ dt } = a \left[ \frac{ 4t }{ (1-t^2)^2 } \right] = \frac{4at}{(1-t^2)^2} \)

Next, differentiate equation (2) with respect to \( t \) using the quotient rule:
\( \frac { dy }{ dt } = b \left[ \frac{ (1-t^2) \frac{d}{dt}(2t) - (2t) \frac{d}{dt}(1-t^2) }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dy }{ dt } = b \left[ \frac{ (1-t^2)(2) - (2t)(-2t) }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dy }{ dt } = b \left[ \frac{ 2 - 2t^2 + 4t^2 }{ (1-t^2)^2 } \right] \)
\( \implies \frac { dy }{ dt } = b \left[ \frac{ 2 + 2t^2 }{ (1-t^2)^2 } \right] = \frac{2b(1+t^2)}{(1-t^2)^2} \)

Finally, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } = \frac { \frac{2b(1+t^2)}{(1-t^2)^2} }{ \frac{4at}{(1-t^2)^2} } \)
\( \implies \frac { dy }{ dx } = \frac{2b(1+t^2)}{4at} \)
\( \implies \frac { dy }{ dx } = \frac{b(1+t^2)}{2at} \)
In simple words: We find the derivative of x and y with respect to t separately, using the division rule for differentiation. After that, we divide the derivative of y by the derivative of x to get the final answer. This helps us see the direct relationship between y and x.

🎯 Exam Tip: The quotient rule \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \) is crucial for these types of problems. Simplify intermediate steps carefully to avoid errors in the final division.

Question 4. If x = \( a(\theta + \sin \theta) \) and y = \( a(1 - \cos \theta) \).
Answer: We are given the parametric equations:
\( x = a(\theta + \sin \theta) \) ...(1)
\( y = a(1 - \cos \theta) \) ...(2)

First, differentiate equation (1) with respect to \( \theta \):
\( \frac { dx }{ d\theta } = a \frac { d }{ d\theta }(\theta + \sin \theta) \)
\( \implies \frac { dx }{ d\theta } = a(1 + \cos \theta) \)

Next, differentiate equation (2) with respect to \( \theta \):
\( \frac { dy }{ d\theta } = a \frac { d }{ d\theta }(1 - \cos \theta) \)
\( \implies \frac { dy }{ d\theta } = a(0 - (-\sin \theta)) = a \sin \theta \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \)
\( \implies \frac { dy }{ dx } = \frac { a \sin \theta }{ a(1 + \cos \theta) } \)
\( \implies \frac { dy }{ dx } = \frac { \sin \theta }{ 1 + \cos \theta } \)

Using the half-angle trigonometric identities (\( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \) and \( 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \)):
\( \frac { dy }{ dx } = \frac { 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} }{ 2 \cos^2 \frac{\theta}{2} } \)
\( \implies \frac { dy }{ dx } = \frac { \sin \frac{\theta}{2} }{ \cos \frac{\theta}{2} } \)
\( \implies \frac { dy }{ dx } = \tan \frac{\theta}{2} \)
In simple words: We find how both x and y change with respect to the angle \( \theta \). Then, we divide the change in y by the change in x. We use special angle formulas to simplify the final fraction to a tangent function. This process reveals the relationship between the slopes in terms of \( \theta \).

🎯 Exam Tip: Remember your trigonometric identities, especially half-angle formulas for \( \sin \theta \) and \( 1 + \cos \theta \), as they are frequently used to simplify expressions in differentiation problems.

Question 5. If x = \( b \sin^2 \theta \) and y = \( a \cos^2 \theta \).
Answer: We are given the parametric equations:
\( x = b \sin^2 \theta \) ...(1)
\( y = a \cos^2 \theta \) ...(2)

First, differentiate equation (1) with respect to \( \theta \) using the chain rule:
\( \frac { dx }{ d\theta } = b \cdot 2 \sin \theta \cdot \frac { d }{ d\theta }(\sin \theta) \)
\( \implies \frac { dx }{ d\theta } = 2b \sin \theta \cos \theta \)

Next, differentiate equation (2) with respect to \( \theta \) using the chain rule:
\( \frac { dy }{ d\theta } = a \cdot 2 \cos \theta \cdot \frac { d }{ d\theta }(\cos \theta) \)
\( \implies \frac { dy }{ d\theta } = 2a \cos \theta (-\sin \theta) = -2a \sin \theta \cos \theta \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \)
\( \implies \frac { dy }{ dx } = \frac { -2a \sin \theta \cos \theta }{ 2b \sin \theta \cos \theta } \)
\( \implies \frac { dy }{ dx } = -\frac { a }{ b } \)
In simple words: We first find how x changes with \( \theta \) and how y changes with \( \theta \). After that, we divide the rate of change of y by the rate of change of x. The common parts cancel out, giving a simple fraction as the answer. This constant slope shows a direct linear relationship between y and x.

🎯 Exam Tip: When differentiating \( \sin^n \theta \) or \( \cos^n \theta \), remember to use the chain rule: \( n \sin^{n-1} \theta \cdot \cos \theta \) and \( n \cos^{n-1} \theta \cdot (-\sin \theta) \) respectively. Watch out for the negative sign in the derivative of \( \cos \theta \).

Question 6. If x = \( a \cos^3 t \), y = \( a \sin^3 t \).
Answer: We are given the parametric equations:
\( x = a \cos^3 t \) ...(1)
\( y = a \sin^3 t \) ...(2)

First, differentiate equation (1) with respect to \( t \) using the chain rule:
\( \frac { dx }{ dt } = a \cdot 3 \cos^2 t \cdot \frac { d }{ dt }(\cos t) \)
\( \implies \frac { dx }{ dt } = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t \)

Next, differentiate equation (2) with respect to \( t \) using the chain rule:
\( \frac { dy }{ dt } = a \cdot 3 \sin^2 t \cdot \frac { d }{ dt }(\sin t) \)
\( \implies \frac { dy }{ dt } = 3a \sin^2 t \cos t \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { 3a \sin^2 t \cos t }{ -3a \cos^2 t \sin t } \)
\( \implies \frac { dy }{ dx } = -\frac { \sin t }{ \cos t } \)
\( \implies \frac { dy }{ dx } = -\tan t \)
In simple words: We calculate how x and y change with respect to t separately. Then, we divide the change in y by the change in x. After simplifying the fraction, we get a negative tangent function. This shows the slope of the curve formed by these parametric equations.

🎯 Exam Tip: When dealing with powers of trigonometric functions like \( \cos^3 t \), always remember the chain rule: differentiate the power first, then differentiate the inner trigonometric function. Carefully cancel common terms to simplify the final expression.

Question 7. x = a(cos t + t sin t), y = a(sin t – t cos t).
Answer: We are given the parametric equations:
\( x = a(\cos t + t \sin t) \) ...(1)
\( y = a(\sin t - t \cos t) \) ...(2)

First, differentiate equation (1) with respect to \( t \). Remember to use the product rule for \( t \sin t \):
\( \frac { dx }{ dt } = a \left[ \frac { d }{ dt }(\cos t) + \frac { d }{ dt }(t \sin t) \right] \)
\( \implies \frac { dx }{ dt } = a [-\sin t + (1 \cdot \sin t + t \cdot \cos t)] \)
\( \implies \frac { dx }{ dt } = a [-\sin t + \sin t + t \cos t] \)
\( \implies \frac { dx }{ dt } = a(t \cos t) = at \cos t \)

Next, differentiate equation (2) with respect to \( t \). Remember to use the product rule for \( t \cos t \):
\( \frac { dy }{ dt } = a \left[ \frac { d }{ dt }(\sin t) - \frac { d }{ dt }(t \cos t) \right] \)
\( \implies \frac { dy }{ dt } = a [\cos t - (1 \cdot \cos t + t \cdot (-\sin t))] \)
\( \implies \frac { dy }{ dt } = a [\cos t - (\cos t - t \sin t)] \)
\( \implies \frac { dy }{ dt } = a [\cos t - \cos t + t \sin t] \)
\( \implies \frac { dy }{ dt } = a(t \sin t) = at \sin t \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { at \sin t }{ at \cos t } \)
\( \implies \frac { dy }{ dx } = \frac { \sin t }{ \cos t } \)
\( \implies \frac { dy }{ dx } = \tan t \)
In simple words: We find the change in x and y with respect to t, using the product rule where needed. Then, we divide the change in y by the change in x. Many terms cancel out, leaving a simple tangent function as the slope. This is how the position of a point changes along the curve.

🎯 Exam Tip: For expressions involving products like \( t \sin t \) or \( t \cos t \), always apply the product rule \( \frac{d}{dt}(uv) = u'v + uv' \). Careful handling of signs, especially after a subtraction, is vital to avoid errors.

Question 8. x = \( 3 \cos t - 2 \cos^3 t \), y = \( 3 \sin t - 2 \sin^3 t \).
Answer: We are given the parametric equations:
\( x = 3 \cos t - 2 \cos^3 t \) ...(1)
\( y = 3 \sin t - 2 \sin^3 t \) ...(2)

First, differentiate equation (1) with respect to \( t \):
\( \frac { dx }{ dt } = \frac { d }{ dt }(3 \cos t) - \frac { d }{ dt }(2 \cos^3 t) \)
\( \implies \frac { dx }{ dt } = -3 \sin t - 2 \cdot 3 \cos^2 t (-\sin t) \)
\( \implies \frac { dx }{ dt } = -3 \sin t + 6 \cos^2 t \sin t \)
\( \implies \frac { dx }{ dt } = 3 \sin t (2 \cos^2 t - 1) \)

Next, differentiate equation (2) with respect to \( t \):
\( \frac { dy }{ dt } = \frac { d }{ dt }(3 \sin t) - \frac { d }{ dt }(2 \sin^3 t) \)
\( \implies \frac { dy }{ dt } = 3 \cos t - 2 \cdot 3 \sin^2 t (\cos t) \)
\( \implies \frac { dy }{ dt } = 3 \cos t - 6 \sin^2 t \cos t \)
\( \implies \frac { dy }{ dt } = 3 \cos t (1 - 2 \sin^2 t) \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { 3 \cos t (1 - 2 \sin^2 t) }{ 3 \sin t (2 \cos^2 t - 1) } \)
\( \implies \frac { dy }{ dx } = \frac { \cos t (1 - 2 \sin^2 t) }{ \sin t (2 \cos^2 t - 1) } \)

Using the double angle identity \( \cos 2t = 1 - 2 \sin^2 t \) and \( \cos 2t = 2 \cos^2 t - 1 \):
\( \frac { dy }{ dx } = \frac { \cos t (\cos 2t) }{ \sin t (\cos 2t) } \)
\( \implies \frac { dy }{ dx } = \frac { \cos t }{ \sin t } \)
\( \implies \frac { dy }{ dx } = \cot t \)
In simple words: We find how x and y change with respect to t by differentiating each part. Then, we divide the change in y by the change in x and simplify the expression using known trigonometric rules for double angles. This leads to a concise cotangent function as the final slope.

🎯 Exam Tip: Recognizing double angle identities for cosine (e.g., \( \cos 2t = 1 - 2 \sin^2 t = 2 \cos^2 t - 1 \)) is crucial for simplifying the final expression. Factor out common terms before cancelling.

Question 9. Find \( \frac { dy }{ dx } \) when x = log t, y = sin t.
Answer: We are given the parametric equations:
\( x = \log t \) ...(1)
\( y = \sin t \) ...(2)

First, differentiate equation (1) with respect to \( t \):
\( \frac { dx }{ dt } = \frac { d }{ dt }(\log t) = \frac { 1 }{ t } \)

Next, differentiate equation (2) with respect to \( t \):
\( \frac { dy }{ dt } = \frac { d }{ dt }(\sin t) = \cos t \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\( \implies \frac { dy }{ dx } = \frac { \cos t }{ \frac { 1 }{ t } } \)
\( \implies \frac { dy }{ dx } = t \cos t \)
In simple words: First, find the derivative of x with respect to t and the derivative of y with respect to t. Then, divide the two derivatives. This gives us the rate at which y changes when x changes, expressed in terms of t.

🎯 Exam Tip: Remember the standard derivatives of common functions: \( \frac{d}{dt}(\log t) = \frac{1}{t} \) and \( \frac{d}{dt}(\sin t) = \cos t \). Be careful when dividing by a fraction; it's equivalent to multiplying by its reciprocal.

Question 10. If x = a (cos \( \theta \) + log tan \( \frac { \theta }{ 2 } \)) and y = a sin \( \theta \), find \( \frac { dy }{ dx } \) at \( \theta = \frac { \pi }{ 4 } \).
Answer: We are given the parametric equations:
\( x = a \left( \cos \theta + \log \tan \frac{\theta}{2} \right) \) ...(1)
\( y = a \sin \theta \) ...(2)

First, differentiate equation (1) with respect to \( \theta \):
\( \frac { dx }{ d\theta } = a \left[ \frac { d }{ d\theta }(\cos \theta) + \frac { d }{ d\theta }\left(\log \tan \frac{\theta}{2}\right) \right] \)
\( \implies \frac { dx }{ d\theta } = a \left[ -\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \sec^2 \frac{\theta}{2} \cdot \frac{1}{2} \right] \)
\( \implies \frac { dx }{ d\theta } = a \left[ -\sin \theta + \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \cdot \frac{1}{\cos^2 \frac{\theta}{2}} \cdot \frac{1}{2} \right] \)
\( \implies \frac { dx }{ d\theta } = a \left[ -\sin \theta + \frac{1}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \right] \)
Using the identity \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \):
\( \implies \frac { dx }{ d\theta } = a \left[ -\sin \theta + \frac{1}{\sin \theta} \right] \)
\( \implies \frac { dx }{ d\theta } = a \left[ \frac{-\sin^2 \theta + 1}{\sin \theta} \right] \)
\( \implies \frac { dx }{ d\theta } = a \frac{\cos^2 \theta}{\sin \theta} \)

Next, differentiate equation (2) with respect to \( \theta \):
\( \frac { dy }{ d\theta } = a \frac { d }{ d\theta }(\sin \theta) = a \cos \theta \)

Now, find \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \)
\( \implies \frac { dy }{ dx } = \frac { a \cos \theta }{ a \frac{\cos^2 \theta}{\sin \theta} } \)
\( \implies \frac { dy }{ dx } = \frac { \cos \theta \cdot \sin \theta }{ \cos^2 \theta } \)
\( \implies \frac { dy }{ dx } = \frac { \sin \theta }{ \cos \theta } = \tan \theta \)

Finally, evaluate \( \frac { dy }{ dx } \) at \( \theta = \frac { \pi }{ 4 } \):
At \( \theta = \frac { \pi }{ 4 } \), \( \frac { dy }{ dx } = \tan \frac { \pi }{ 4 } = 1 \).
(If asked for \( \theta = \frac { \pi }{ 3 } \), then \( \frac { dy }{ dx } = \tan \frac { \pi }{ 3 } = \sqrt{3} \).)
In simple words: First, we find how both x and y change with respect to \( \theta \), simplifying x's derivative using trigonometric identities. Then, we divide the change in y by the change in x to get \( \tan \theta \). Lastly, we put the given value of \( \theta \) into our answer to find the specific slope at that point. This shows the slope of the curve at a particular angle.

🎯 Exam Tip: For problems involving \( \log(\tan(\frac{\theta}{2})) \), remember to apply the chain rule multiple times: differentiate log, then tan, then \( \frac{\theta}{2} \). Also, recall the identity \( \frac{1}{\tan \alpha} \cdot \sec^2 \alpha = \frac{\cos \alpha}{\sin \alpha} \cdot \frac{1}{\cos^2 \alpha} = \frac{1}{\sin \alpha \cos \alpha} = \frac{2}{\sin(2\alpha)} \).