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U n s a f e C o d e , P o i n t e r s , N u l l a b l e Ty p e s , a n d M i s c e l l a n e o u s To p i c s

Solution Let the q s and b s be the angles for each portion shown Note that for velocity and acceleration one should multiply values by w and w2, respectively, and the bound ary conditions are y(0) = 0, y(0) = 0, y(b4) = 0, and y(b4) = total rise h Use basic trigonometric relationships for

x 2 (x 4 + 4x 2 + 5)1/2

if(resultHasValue) ConsoleWriteLine("result has this value: " + resultValue); else ConsoleWriteLine("result has no value"); } }

The output is shown here:

Portion I y = A sin(pq 2b1 ) y = (2b1 A p )[1 - cos(pq 2b1 )] y = - A(2b1 p ) sin(pq 2b1 ) + (2 Ab1 p )q y1 = 2 Ab1 p y1 = (2 Ab12 p )(1 - 2 p )

result has no value result has this value: 110

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If you attempt to use a cast to convert a nullable object to its underlying type, a SystemInvalidOperationException will be thrown if the nullable object contains a null value This can occur, for example, when you use a cast to assign the value of a nullable object to a variable of its underlying type You can avoid the possibility of this exception being thrown by using the operator, which is called the null coalescing operator It lets you specify a default value that will be used when the nullable object contains null It also eliminates the need for the cast The operator has this general form: nullable-object default-value If nullable-object contains a value, then the value of the is that value Otherwise, the value of the operation is default-value For example, in the following code balance is null This causes currentBalance to be assigned the value 00 and no exception will be thrown

SOLUTION The limit as written is of the form We rewrite it as 1 (1 + 4x 2 + 5x 4 )1/2 x + x + x 2 Notice that both the numerator and denominator tend to zero, so it is now indeterminate of the form 0/0 We may thus apply l H pital s Rule The result is that the limit equals lim x 2 1 (1 + 4x 2 + 5x 4 )1/2 = lim ( 1/2)(1 + 4x 2 + 5x 4 ) 1/2 ( 8x 3 20x 5 ) x + 2x 3 lim = lim (1 + 4x 2 + 5x 4 ) 1/2 (2 + 5x 2 )

2 y = - A(b3 p ) [1 - cos(pq b3 )] + y2q + y2 y3 = y2 2 y3 = 2 A(b3 p ) + Ab 2 b3 + 2 Ab1b3 p + Ab 2 2 + 2 Ab1b 2 p + (2 Ab12 p )(1 - 2 p ) 2

double balance = null; double currentBalance; currentBalance = balance 00;

In the next sequence, balance is given the value 12375:

Portion IV y = -A y = - Aq 4 + y3 y = -[ A(q 4 ) 2] + y3q + y3 y4 = - Ab 2 + y3 = - Ab 4 + Ab 2 + 2 Ab1 p = 0 b 4 = b 2 + 2 b1 p

double balance = 12375; double currentBalance; currentBalance = balance 00;

x 2 (x 4 + 4x 2 + 5)1/2 = 2

Now, currentBalance will contain the value of balance, which is 12375 One other point: The right-hand expression of the is evaluated only if the left-hand expression does not contain a value The following program demonstrates this fact:

h 2 2 2 + b 2 b 4 + 2b1b 4 p + 2(b3 p ) + b2 b3 + 2b1b3 p + b2 2 + 2b1b2 p + (2b12 p )(1 - 2 p )]

// Using using System; class NullableDemo2 {

Part I:

For a given total rise h for angle q0 and any two of the b angles given one can solve for all angles and all values of the derivative curves

x

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