Question
A particle moves along the curve x^2+4=y. The point on the curve at which the y coordinate changes twice as fast as the x coordinate,is
Answers
you want dy/dt = 2 dx/dt
x^2+4 = y
2x dx/dt = dy/dt
So, you want 2x dx/dt = 2 dx/dt
2x = 2
x = 1
So, at (1,5) dy/dt = 2 dx/dt
or, you want dy/dx = 2
y = x^2+4
dy/dx = 2x = 2
x = 1
x^2+4 = y
2x dx/dt = dy/dt
So, you want 2x dx/dt = 2 dx/dt
2x = 2
x = 1
So, at (1,5) dy/dt = 2 dx/dt
or, you want dy/dx = 2
y = x^2+4
dy/dx = 2x = 2
x = 1
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