Graph the image of the given triangle under a dilation with a scale factor of -3 and center of dilation (0,0).
2 answers:
<h2>Answer with explanation:</h2>
We are given a triangle with vertices:
Let A(-3,-3) , B(0,2) and C(-2,3)
Now, the triangle is to be dilated with a scale factor of -3 and center of dilation (0,0).
This means that each of the vertices of the triangle will be transformed by using the rule:
(x,y) → (-3x,-3y)
Hence, we get:
A(-3,-3) → A'(9,9)
B(0,2) → B'(0,-6)
C(-2,3) → C'(6,-9)
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Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
Then r' = (1, 2t, 3t^2)
General Line integral is:
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
After integrating using power rule, replace 'u' with function for 't' and evaluate limits: