I can't seem to get this question right:
If x=2/t and y=3t^2-1, express dy/dx in terms of t. Evaluate dy/dx at the point (2,2)
I found dy/dx to be -12t^3 but what do i do with the point now?
Thanks in advance:)
If x=2/t and y=3t^2-1, express dy/dx in terms of t. Evaluate dy/dx at the point (2,2)
I found dy/dx to be -12t^3 but what do i do with the point now?
Thanks in advance:)
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dx/dt = -2/t²
dy/dt = 6t
dy/dx = dy/dt ÷ dx/dt
dy/dx = 6t ÷ -2/t²
dy/dx = -6t³/2
dy/dx = -3t³
At (2, 2), we know x = 2 so using x = 2/t we see that t = 1.
When t = 1, dy/dx = -3 x 1³ = -3
Alternative method for the second part of the question
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t = 2/x
Substitute this into y = 3t² - 1
y = 3(2/x)² - 1
y = 12/x² - 1
dy/dx = -24/x³
When x = 2, dy/dx = -24/2³ = -24/8 = -3
dy/dt = 6t
dy/dx = dy/dt ÷ dx/dt
dy/dx = 6t ÷ -2/t²
dy/dx = -6t³/2
dy/dx = -3t³
At (2, 2), we know x = 2 so using x = 2/t we see that t = 1.
When t = 1, dy/dx = -3 x 1³ = -3
Alternative method for the second part of the question
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
t = 2/x
Substitute this into y = 3t² - 1
y = 3(2/x)² - 1
y = 12/x² - 1
dy/dx = -24/x³
When x = 2, dy/dx = -24/2³ = -24/8 = -3