. A large tank is partially filled with a solution. The tank has a faucet that allows solution to enter the tank at a rate of 3
1 answer:
You might be interested in
Answer:
( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 ), ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )
y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2
y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1
Step-by-step explanation:
Given:
- Four functions to construct a diamond:
y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, and y = x − √ 2.
Find:
a)Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.
b)Find the intersection points between the unit circle and each of the four lines.
(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.
Solution:
- For first part see the attachment.
- The equation of the unit circle is given as follows:
x^2 + y^2 = 1
- To determine points of intersection we have to solve each given function of y with unit circle equation for set of points of intersection:
For: y = −x+ √ 2 , x - √ 2
And: x^2 + y^2 = 1
x^2 + (+/- * (x - √ 2))^2 = 1
x^2 + (x - √ 2)^2 = 1
2x^2 -2√ 2*x + 2 = 1
2x^2 -2√ 2*x + 1 = 0
2[ x^2 - √ 2] + 1 = 0
Complete sqr: (1 - 1/√ 2)^2 = 0
x = 1/√ 2 , x = 1/√ 2
y = -1/√ 2 + √ 2 = 1/√ 2
y = 1/√ 2 - √ 2 = - 1/√ 2
Points are: ( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 )
- Using vertical symmetry of unit circle we can also evaluate other intersection points by intuition:
x = - 1/√ 2
y = 1/√ 2 , -1/√ 2
Points are: ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )
- To determine the function for the rhombus region that would be tangential to unit circle with center at ( - 2 , - 1 ):
- To shift our unit circle from origin to ( - 2 , - 1 ) i.e two units left and 1 unit down.
- For shifts we use the following substitutions:
x = x + 2 ....... 2 units of left shift
y = y + 1 .......... 1 unit of down shift
- Now substitute the above shifting expression in all for functions we have:
y = −x+ √ 2 -----> y + 1 = - ( x + 2 ) + √ 2
y = −x− √ 2 -----> y + 1 = - ( x + 2 ) - √ 2
y = x- √ 2 -------> y + 1 = ( x + 2 ) - √ 2
y = x+ √ 2 ------> y + 1 = ( x + 2 ) + √ 2
x^2 + y^2 = 1 -----> ( x + 2 )^2 + ( y + 1)^2 = 1
- The following diamond shape graph would have the 4 functions as:
y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 , y + 1 = ( x + 2 ) - √ 2
y + 1 = ( x + 2 ) + √ 2 , ( x + 2 )^2 + ( y + 1)^2 = 1
- See attachment for the new sketch.
