Asked by Anonymous
Find dy/dx at the point (-3, 1) for the equation x = y^3-9y^2+5.
A. -1/17
B. -1/13
C. -1/15
D. 1/17
E. 1/15
I posted this question before with a typo, but the point still doesn't appear to be on the graph of the function.
A. -1/17
B. -1/13
C. -1/15
D. 1/17
E. 1/15
I posted this question before with a typo, but the point still doesn't appear to be on the graph of the function.
Answers
Answered by
Damon
yes it is. Perhaps you are misled by the x being the independent variable instead of y
When y = 1
x = 1^3 = 9 (1)^2 + 5
= 1 - 9 + 5
= -3 sure enough
now the problem
dx/dx = 1 = 3 y^2 dy/dx - 18y dy/dx
1 = (3y^2-18y)dy/dx
dy/dx = 1/ (3*1^2 -18*1)
dy/dx = 1/(-15) = -1/15
amazing - that is one of the choices !
When y = 1
x = 1^3 = 9 (1)^2 + 5
= 1 - 9 + 5
= -3 sure enough
now the problem
dx/dx = 1 = 3 y^2 dy/dx - 18y dy/dx
1 = (3y^2-18y)dy/dx
dy/dx = 1/ (3*1^2 -18*1)
dy/dx = 1/(-15) = -1/15
amazing - that is one of the choices !
Answered by
Damon
Sometimes conventional obscurity is insufficient and extraordinary efforts must be made to confuse the student. One seemingly simple but surprisingly effective technique is to reverse the usual roles of x and y.
Answered by
Anonymous
Haha! Thank you!
Answered by
Damon
You are welcome.
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