(a) Answer The required value of d y d x is 2 t + 1 2 t − 1 . Explanation Given information: The parametric equations: x = t 2 + t , y = t 2 − t Calculation: The parametric equations are x = t 2 + t , y = t 2 − t . Differentiate the above equations with respect to t . d x d t = d d t [ t 2 + t ] = 2 t + 1 And, d y d t = d d t [ t 2 − t ] = 2 t − 1 Now, d y d x = d y d t d x d t = 2 t + 1 2 t − 1 Hence, the required value of d y d x is 2 t + 1 2 t − 1 . (b) To determine To calculate: The value of d 2 y d x 2 . Answer The required value of d 2 y d x 2 is 4 ( 2 t + 1 ) 3 . Explanation Given information: The parametric equations: x = t 2 + t , y = t 2 − t Calculation: The parametric equations are x = t 2 + t , y = t 2 − t . Differentiate the above equations with respect to t . d x d t = d d t [ t 2 + t ] = 2 t + 1 And, d y d t = d d t [ t 2 − t ] = 2 t − 1 Now, d y d x = y ′ = d y d t d x d t = 2 t + 1 2 t − 1 Again, d 2 y d x 2 = ( d y ′ d t d x d t ) = [ ( 2 t + 1 ) ( 2 ) − ( 2 t − 1 ) ( 2 ) ] ( 2 t + 1 ) 2 ( 2 t + 1 ) = 4 t + 2 − 4 t + 2 ( 2 t + 1 ) 3 = 4 ( 2 t + 1 ) 3 Hence, the required value of d 2 y d x 2 is 4 ( 2 t + 1 ) 3 .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 11.1, Problem 12E

(a)

To determine

To calculate: The value of $dydx$ .

(a)

Expert Solution

Answer to Problem 12E

The required value of $dydx$ is $2t+12t−1$ .

Explanation of Solution

Given information:

The parametric equations: $x=t2+t$ , $y=t2−t$

Calculation:

The parametric equations are $x=t2+t$ , $y=t2−t$ .

Differentiate the above equations with respect to t .

$dxdt=ddt[t2+t]=2t+1$

And, $dydt=ddt[t2−t]=2t−1$

Now, $dydx=dydtdxdt=2t+12t−1$

Hence, the required value of $dydx$ is $2t+12t−1$ .

(b)

To determine

To calculate: The value of $d2ydx2$ .

(b)

Expert Solution

Answer to Problem 12E

The required value of $d2ydx2$ is $4(2t+1)3$ .

Explanation of Solution

Given information:

The parametric equations: $x=t2+t$ , $y=t2−t$

Calculation:

The parametric equations are $x=t2+t$ , $y=t2−t$ .

Differentiate the above equations with respect to t .

$dxdt=ddt[t2+t]=2t+1$

And, $dydt=ddt[t2−t]=2t−1$

Now, $dydx=y′=dydtdxdt=2t+12t−1$

Again,

$d2ydx2=(dy′dtdxdt)=[(2t+1)(2)−(2t−1)(2)](2t+1)2(2t+1)=4t+2−4t+2(2t+1)3=4(2t+1)3$

Hence, the required value of $d2ydx2$ is $4(2t+1)3$ .

Chapter 11.1, Problem 11E

Chapter 11.1, Problem 13E

Chapter 11 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 11.1 - Prob. 1QR Ch. 11.1 - Prob. 2QR Ch. 11.1 - Prob. 3QR Ch. 11.1 - Prob. 4QR Ch. 11.1 - Prob. 5QR Ch. 11.1 - Prob. 6QR Ch. 11.1 - Prob. 7QR Ch. 11.1 - Prob. 8QR Ch. 11.1 - Prob. 9QR Ch. 11.1 - Prob. 10QR

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