a curve on the xy-plane is defined parametically by the equations x=t^2 +t and y=t^2-t.?

Question

a curve on the xy-plane is defined parametically by the equations x=t^2 +t and y=t^2-t. For what values of t is the tangent to the curve horizontal?

Answer 1

I obviously don't know the easy way to do this,
so here's what I did -
x = t^2 + t and y = t^2 - t

Find the value of t in both equations :

t^2 + t - x = 0 implies t = [-1 ± sqrt(4x + 1)] / 2
and
t^2 - t - y = 0 implies t = [1 ± sqrt(4y + 1)] / 2

Setting these equal to each other gives :
y = x + 1 ± sqrt(4x + 1)

Looking at the graphs of these, it is seen that the positive
valued graph does not possess a horizontal tangent.
Thus, the equation is :
y = x + 1 - sqrt(4x + 1)

Taking the derivative gives :
y' = 1 - 2/sqrt(4x + 1)

Setting equal to zero (for horizontal tangent) gives :
x = 3/4

Plugging this back into t = [-1 ± sqrt(4x + 1)] / 2 gives :
t = -3/2 or 1/2

Answer 2

actually, you have an error in your calculations. the exponent of T doesnt relate with the -t in y=t^2-t. Therefore, the parallel equation would simply be a parabola colliding with its so called parallel relation. if you were to modify this in some way and tweak it til it works, the curve horizontal would end up being AT LEAST 1/99. Otherwise it would be a negative slope and a perpendicular parabola wouldnt relate, causing an impossible equation.

Answer 3

1 & -1

Answer 4

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Answer 5

z = 1/2 root(x^2+y^2)

@z/@x = 0 = x/2(x^2+y^2)^(-1/2)


thus x = 0 = 2t^2+t, then ,
2t = -1 , t = -1/2

Answer 6

dy/dx = (2t-1)/(2t+1)
dy/dx = 0 when t = 1/2

Answer 7

(2^t-1)(2^t+1) = (2^t)(4t^2)=1/2