Rewrite the following parametric equations by solving for y
x(t) = e^-t
y(t) = 3e^2t
a) y=1/3x^2,x>0
b)3/x^2, x>0
c) y= 3e^t, x>0
d) y=4e^t, x>0
x(t) = e^-t
ln x = - t
t = - ln x
y = 3 e ^ ( 2 t )
y = 3 e ^ [ 2 ∙ ( - ln x ) ]
y = 3 e ^ [ - ( 2 ∙ ln x ) ]
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Remark:
e ^ [ ( 2 ∙ ln x ) ] = x ^ 2
e ^ [ ( - 2 ∙ ln x ) ] = 1 / x ^ 2
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So:
y = 3 e ^ [ - ( 2 ∙ ln x ) ]
y = 3 / x ^ 2
x > 0
x(t) = e^-t
x^-2 = (e^-t)^-2 = e^2t
3 x^-2 = 3 e^2t = 3/x^2 = y
To rewrite the given parametric equations by solving for y, we will substitute the given x(t) values into the equation for y(t). Let's solve each case:
a) y = 1/3x^2, x > 0
Substituting x(t) = e^-t:
y(t) = 1/3(e^-t)^2
y(t) = 1/3e^-2t
b) y = 3/x^2, x > 0
Substituting x(t) = e^-t:
y(t) = 3/(e^-t)^2
y(t) = 3/e^-2t
y(t) = 3e^2t
c) y = 3e^t, x > 0
There is no need to solve for y in this case since the given y(t) equation is already in terms of e^t. It matches the original equation.
d) y = 4e^t, x > 0
Again, the given y(t) equation is already in terms of e^t, so no need to solve for y. It matches the original equation.
To rewrite the parametric equations by solving for y, we need to eliminate the parameter t and express y in terms of x.
a) For the equation y = 1/3x^2, x > 0:
Since y is already expressed in terms of x, there is no need to eliminate the parameter t. Therefore, the rewritten parametric equations are:
x(t) = e^-t
y(t) = 1/3(e^-t)^2
b) For the equation y = 3/x^2, x > 0:
To eliminate the parameter t, we need to substitute the expression for x(t) = e^-t in terms of x, and then solve for y:
x(t) = e^-t --> e^t = x --> t = ln(x)
Substituting the value of t = ln(x) into y(t), we get:
y(t) = 3(e^2t) --> y = 3e^2(ln(x))
Thus, the rewritten parametric equations are:
x(t) = e^-t
y(t) = 3e^2(ln(x))
c) For the equation y = 3e^t, x > 0:
Similarly to the previous question, we substitute the expression for x(t) = e^-t in terms of x, and then solve for y:
x(t) = e^-t --> e^t = x --> t = ln(x)
Substituting the value of t = ln(x) into y(t), we get:
y(t) = 3(e^t) --> y = 3e^(ln(x))
Since e^(ln(x)) is equivalent to x, the rewritten parametric equations are:
x(t) = e^-t
y(t) = 3x
d) For the equation y = 4e^t, x > 0:
Using the same process as before, we substitute the expression for x(t) = e^-t in terms of x, and then solve for y:
x(t) = e^-t --> e^t = x --> t = ln(x)
Substituting the value of t = ln(x) into y(t), we get:
y(t) = 4(e^t) --> y = 4e^(ln(x))
Since e^(ln(x)) is equivalent to x, the rewritten parametric equations are:
x(t) = e^-t
y(t) = 4x