The parametric equations of a curve are x = t + 4 ln t , y = t + 9 t for t > 0 . Show that d y d x = t 2 - 9 t 2 + 4 t .

Given, the parametric equations of a curve are x = t + 4 ln t , , and y = t + 9 t for t > 0 . To Prove : d y d x = t 2 - 9 t 2 + 4 t We know that when a curve is given in parametric form, in terms of the parameter t we can use the chain rule to find d y d x in terms of t . Since x = t + 4 ln t ⇒ d x d t = d d t t + 4 ln t = d d t t + 4 d d t ln t = 1 + 4 t . . . . i ∵ d d t ln t = 1 t And, y = t + 9 t ⇒ d y d t = d d t t + 9 t - 1 = 1 + 9 × - t - 2 ∵ d d t t n = n t n - 1 , n ∈ R = 1 - 9 t 2 . . . . . i i Now, applying the chain rule, we have d y d x = d y d t × d t d x = 1 - 9 t 2 × t t + 4 [Using i , and i i ] = t 2 - 9 t t + 4 Hence, d y d x = t 2 - 9 t 2 + 4 t .

The parametric equations of a curve are x = 2 sin θ + cos 2 θ , y = 1 + cos 2 θ , for 0 ⩽ θ ⩽ π 2 . Show that d y d x = 2 sin θ 2 sin θ - 1 .

Given, the parametric equations of a curve are x = 2 sin θ + cos 2 θ , and y = 1 + cos 2 θ , To Prove: d y d x = 2 sin θ 2 sin θ - 1 We know that when a curve is given in parametric form, in terms of the parameter θ we can use the chain rule to find d y d x in terms of θ . Since x = 2 sin θ + cos 2 θ ⇒ d x d θ = d d θ 2 sin θ + cos 2 θ = 2 d d θ sin θ + d d θ cos 2 θ = 2 cos θ + - sin 2 θ × 2 ∵ d d x sin x = cos x , and d d x cos x = - sin x = 2 cos θ - 2 sin 2 θ . . . . i And, y = 1 + cos 2 θ ⇒ d y d θ = d d θ 1 + cos 2 θ = d d θ 1 + d d θ cos 2 θ = 0 + - sin 2 θ × 2 = - 2 sin 2 θ . . . . . i i Now, applying the chain rule, we have d y d x = d y d θ × d θ d x = - 2 sin 2 θ × 1 2 cos θ - 2 sin 2 θ [Using i , and i i ] = - sin 2 θ cos θ - sin 2 θ = - 2 sin θ cos θ cos θ - 2 sin θ cos θ ∵ sin 2 A = 2 sin A cos A = 2 sin θ 2 sin θ - 1 Hence, d y d x = 2 sin θ 2 sin θ - 1

The parametric equations of a curve are x = 2 sin θ + cos 2 θ , y = 1 + cos 2 θ , for 0 ⩽ θ ⩽ π 2 . Find the coordinates of the point on the curve where the tangent is parallel to the x -axis.

Given, the parametric equations of a curve are x = 2 sin θ + cos 2 θ , and y = 1 + cos 2 θ , We need to find the coordinates of the point on the curve where the tangent is parallel to the x -axis. We know that when a curve is given in parametric form, in terms of the parameter θ we can use the chain rule to find d y d x in terms of θ . Since x = 2 sin θ + cos 2 θ ⇒ d x d θ = d d θ 2 sin θ + cos 2 θ = 2 d d θ sin θ + d d θ cos 2 θ = 2 cos θ + - sin 2 θ × 2 ∵ d d x sin x = cos x , and d d x cos x = - sin x = 2 cos θ - 2 sin 2 θ . . . . i And, y = 1 + cos 2 θ ⇒ d y d θ = d d θ 1 + cos 2 θ = d d θ 1 + d d θ cos 2 θ = 0 + - sin 2 θ × 2 = - 2 sin 2 θ . . . . . i i Now, applying the chain rule, we have d y d x = d y d θ × d θ d x = - 2 sin 2 θ × 1 2 cos θ - 2 sin 2 θ [Using i , and i i ] = - sin 2 θ cos θ - sin 2 θ = - 2 sin θ cos θ cos θ - 2 sin θ cos θ ∵ sin 2 A = 2 sin A cos A = 2 sin θ 2 sin θ - 1 Hence, d y d x = 2 sin θ 2 sin θ - 1 . . . . i i i We know that d y d x is the gradient of the tangent to the given curve. Also, we know that any line is parallel to the x -axis if the gradient of that line is zero. Thus, for the tangent to be parallel to x - axis, we must have d y d x = 0 ⇒ 2 sin θ 2 sin θ - 1 = 0 [using i i i ] ⇒ sin θ = 0 ⇒ θ = n π , n ∈ Z But the curve is given for 0 ⩽ θ ⩽ π 2 . Thus, the tangent to the given curve is parallel to x -axis where θ = 0 . Now, at θ = 0 x = 2 sin θ = 2 sin 0 = 0 , and y = 1 + cos 2 θ = 1 + cos 0 = 1 + 1 = 2 Hence, the required coordinates of the point on the curve where the tangent is parallel to the x -axis are 1 , 2 .

The parametric equations of a curve are x = 2 sin θ + cos 2 θ , y = 1 + cos 2 θ , for 0 ⩽ θ ⩽ π 2 . Show that the tangent to the curve at the point 3 2 , 3 2 is parallel to the y -axis.

Given, the parametric equations of a curve are x = 2 sin θ + cos 2 θ , and y = 1 + cos 2 θ , We need to find the coordinates of the point on the curve where the tangent is parallel to the x -axis. We know that when a curve is given in parametric form, in terms of the parameter θ we can use the chain rule to find d y d x in terms of θ . Since x = 2 sin θ + cos 2 θ ⇒ d x d θ = d d θ 2 sin θ + cos 2 θ = 2 d d θ sin θ + d d θ cos 2 θ = 2 cos θ + - sin 2 θ × 2 ∵ d d x sin x = cos x , and d d x cos x = - sin x = 2 cos θ - 2 sin 2 θ . . . . i And, y = 1 + cos 2 θ ⇒ d y d θ = d d θ 1 + cos 2 θ = d d θ 1 + d d θ cos 2 θ = 0 + - sin 2 θ × 2 = - 2 sin 2 θ . . . . . i i Now, applying the chain rule, we have d y d x = d y d θ × d θ d x = - 2 sin 2 θ × 1 2 cos θ - 2 sin 2 θ [Using i , and i i ] = - sin 2 θ cos θ - sin 2 θ = - 2 sin θ cos θ cos θ - 2 sin θ cos θ ∵ sin 2 A = 2 sin A cos A = 2 sin θ 2 sin θ - 1 Hence, d y d x = 2 sin θ 2 sin θ - 1 . . . . i i i We know that d y d x is the gradient of the tangent to the given curve. Also, we know that any line is parallel to the y -axis if the gradient of that line is not defined. Now, first we need to find the gradient of the tangent to the given curve at the point 3 2 , 3 2 Since y = 1 + cos 2 θ ⇒ 3 2 = 1 + cos 2 θ ⇒ cos 2 θ = 1 2 ⇒ 2 θ = cos - 1 1 2 = π 3 ⇒ θ = π 6 Now, d y d x = 2 sin θ 2 sin θ - 1 d y d x ] θ = π 6 = 2 sin π 6 2 sin π 6 - 1 = 2 × 1 2 2 × 1 2 - 1 = 1 0 = Not Defined. Hence, the tangent to the curve at the point 3 2 , 3 2 is parallel to the y -axis.

The parametric equations of a curve are x = ln tan t , y = 2 sin 2 t for 0 < t < π 2 . Show that d y d x = sin 4 t .

Given, the parametric equations of a curve are x = ln tan t , and y = 2 sin 2 t To Prove : d y d x = sin 4 t We know that when a curve is given in parametric form, in terms of the parameter t we can use the chain rule to find d y d x in terms of t . Since x = ln tan t ⇒ d x d t = d d t ln tan t = 1 tan t × d d t tan t = sec 2 t tan t . . . . . i ∵ d d t tan t = sec 2 t And, y = 2 sin 2 t ⇒ d y d t = d d t 2 sin 2 t = 2 d d t sin 2 t = 2 cos 2 t × d d t 2 t = 4 cos 2 t . . . . . i i ∵ d d x x n = n x n - 1 Now, applying the chain rule, we have d y d x = d y d t × d t d x = 4 cos 2 t × 1 sec 2 t tan t [Using i , and i i ] = 4 cos 2 t × tan t sec 2 t = 4 cos 2 t × sin t cos t × 1 cos 2 t ∵ tan t = sin t cos t , sec t = 1 cos t = 4 cos 2 t × sin t cos t = 2 cos 2 t × 2 sin t cos t = 2 sin 2 t cos 2 t ∵ sin 2 A = 2 sin A cos A = sin 4 t Hence, d y d x = sin 4 t .

The parametric equations of a curve are x = ln tan t , y = 2 sin 2 t for 0 < t < π 2 . Hence, show that at the point where x = 0 the tangent is parallel to the x -axis.

Given, the parametric equations of a curve are x = ln tan t , and y = 2 sin 2 t We need to show that the tangent is parallel to the x -axis at the point where x = 0 . We know that when a curve is given in parametric form, in terms of the parameter t we can use the chain rule to find d y d x in terms of t . Since x = ln tan t ⇒ d x d t = d d t ln tan t = 1 tan t × d d t tan t = sec 2 t tan t . . . . . i ∵ d d t tan t = sec 2 t And, y = 2 sin 2 t ⇒ d y d t = d d t 2 sin 2 t = 2 d d t sin 2 t = 2 cos 2 t × d d t 2 t = 4 cos 2 t . . . . . i i ∵ d d x x n = n x n - 1 Now, applying the chain rule, we have d y d x = d y d t × d t d x = 4 cos 2 t × 1 sec 2 t tan t [Using i , and i i ] = 4 cos 2 t × tan t sec 2 t = 4 cos 2 t × sin t cos t × 1 cos 2 t ∵ tan t = sin t cos t , sec t = 1 cos t = 4 cos 2 t × sin t cos t = 2 cos 2 t × 2 sin t cos t = 2 sin 2 t cos 2 t ∵ sin 2 A = 2 sin A cos A = sin 4 t Hence, d y d x = sin 4 t . . . . . i i i . Now, when x = 0 x = ln tan t ⇒ ln tan t = 0 ⇒ tan t = 1 ⇒ t = π 4 . Thus, we need to find the gradient of the tangent at t = π 4 d y d x ] t = π 4 = sin 4 π 4 = sinπ = 0 Since the gradient of any line parallel to the x -axis is zero. Hence, the tangent to the given curve at the point where x = 0 is parallel to the x -axis.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G

Attempt the free practice questions on Chapter 4: Differentiation, Exercise 8: EXERCISE 4G with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G with Hints & Solutions

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 8

The parametric equations of a curve are $x = t + 4 \ln t, y = t + \frac{9}{t}$ for $t > 0 .$

Show that $\frac{d y}{d x} = \frac{t^{2} - 9}{t^{2} + 4 t} .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 8

The parametric equations of a curve are $x = t + 4 \ln t, y = t + \frac{9}{t}$ for $t > 0 .$

The curve has one stationary point. Find the $y$ -coordinate of this point and determine whether it is a maximum or a minimum point.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 9

The parametric equations of a curve are $x = 1 + 2 \sin^{2} θ, y = 1 + 2 \tan θ .$ Find the equation of the normal to the curve at the point where $θ = \frac{π}{4} .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 10

The parametric equations of a curve are $x = 2 \sin θ + \cos 2 θ, y = 1 + \cos 2 θ,$ for $0 ⩽ θ ⩽ \frac{π}{2} .$

Show that $\frac{d y}{d x} = \frac{2 \sin θ}{2 \sin θ - 1} .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 10

The parametric equations of a curve are $x = 2 \sin θ + \cos 2 θ, y = 1 + \cos 2 θ,$ for $0 ⩽ θ ⩽ \frac{π}{2} .$

Find the coordinates of the point on the curve where the tangent is parallel to the $x$ -axis.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 10

The parametric equations of a curve are $x = 2 \sin θ + \cos 2 θ, y = 1 + \cos 2 θ,$ for $0 ⩽ θ ⩽ \frac{π}{2} .$

Show that the tangent to the curve at the point $(\frac{3}{2}, \frac{3}{2})$ is parallel to the $y$ -axis.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 11

The parametric equations of a curve are $x = \ln (\tan t), y = 2 \sin 2 t$ for $0 < t < \frac{π}{2} .$

Show that $\frac{d y}{d x} = \sin 4 t .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4G>Q 11

The parametric equations of a curve are $x = \ln (\tan t), y = 2 \sin 2 t$ for $0 < t < \frac{π}{2} .$

Hence, show that at the point where $x = 0$ the tangent is parallel to the $x$ -axis.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 8: EXERCISE 4G with Hints & Solutions

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