a. Answer The value of d y d x = 0. Explanation Given equation y = a x 3 + b x 2 + c x + d . Calculation: The graph y = a x 3 + b x 2 + c x + d y ( 0 ) = 0 d y ( 0 ) d x = 0 d y d x = 0 Thus, the value of d y d x = 0. b. To determine Find d y d x at x = − L . Answer The value of d y ( − L ) d x = 0 . Explanation Given equation y = a x 3 + b x 2 + c x + d . Calculation: Consider the graph and it is given in the question that y ( − L ) = H d y ( − L ) d x = 0 Thus, the value of d y ( − L ) d x = 0 . c. To determine Prove the value y ( x ) = H [ 2 ( x L ) 3 − 3 ( x L ) 2 ] . Answer The value y ( x ) = H [ 2 ( x L ) 3 − 3 ( x L ) 2 ] . Explanation Given equation y = a x 3 + b x 2 + c x + d . Calculation: It is given in the question that y = a x 3 + b x 2 + c x + d and y ( 0 ) = 0. y = a x 3 + b x 2 + c x + d y ( 0 ) = a ( 0 ) 3 + b ( 0 ) 2 + c ( 0 ) + d = 0 d = 0 Also, y = a x 3 + b x 2 + c x + d y ′ = 3 a x 2 + 2 b x + c y ′ ( 0 ) = 0 c = 0 Using these result y ′ ( x ) = 3 a x 2 + 2 b x y ′ ( − L ) = 0 3 a ( − L ) 2 + 2 b ( − L ) = 0 3 a ( L ) 2 = 2 b ( L ) b = 3 a L 2 Substitute this in y ( − L ) = H y ( − L ) = a ( − L ) 3 + b ( − L ) 2 = H = a ( − L ) 3 + 3 a L 2 ( − L ) 2 = H a ( L ) 3 2 = H a = 2 H ( L ) 3 , b = 3 H ( L ) 2 Substitute all the values obtained above then we get, y ( x ) = H [ 2 ( x L ) 3 − 3 ( x L ) 2 ] Thus, the value is y ( x ) = H [ 2 ( x L ) 3 − 3 ( x L ) 2 ] .

Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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Question

Chapter 5.4, Problem 63E

To determine

To find the value of dydx at x=0 .

Expert Solution

Answer to Problem 63E

The value of dydx=0.

Explanation of Solution

Given equation y=ax3+bx2+cx+d.

Calculation:

The graph y=ax3+bx2+cx+d

y(0)=0dy(0)dx=0 dydx=0

Thus, the value of dydx=0.

To determine

Find dydx at x=−L.

Expert Solution

Answer to Problem 63E

The value of dy(−L)dx=0 .

Explanation of Solution

Given equation y=ax3+bx2+cx+d.

Calculation:

Consider the graph and it is given in the question that

y(−L)=Hdy(−L)dx=0

Thus, the value of dy(−L)dx=0 .

To determine

Prove the value y(x)=H[2(xL)3−3(xL)2] .

Expert Solution

Answer to Problem 63E

The value y(x)=H[2(xL)3−3(xL)2] .

Explanation of Solution

Given equation y=ax3+bx2+cx+d.

Calculation:

It is given in the question that

y=ax3+bx2+cx+d and y(0)=0.y=ax3+bx2+cx+dy(0)=a(0)3+b(0)2+c(0)+d=0d=0

Also,

y=ax3+bx2+cx+d y′=3ax2+2bx+c y′(0)=0c=0

Using these result

y′(x)=3ax2+2bx y′(−L)=03a(−L)2+2b(−L)=03a(L)2=2b(L) b=3aL2

Substitute this in

y(−L)=Hy(−L)=a(−L)3+b(−L)2=H =a(−L)3+3aL2(−L)2=Ha(L)32=Ha=2H(L)3,b=3H(L)2

Substitute all the values obtained above then we get,

y(x)=H[2(xL)3−3(xL)2]

Thus, the value is y(x)=H[2(xL)3−3(xL)2] .

Chapter 5.4, Problem 62E

Chapter 5.4, Problem 64E

Chapter 5 Solutions

Calculus: Graphical, Numerical, Algebraic

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