Calculating the Tangent to a Parametric Curve at a Given Point

[53Mark53]

Homework Statement

Consider the parametric curve given by:

x=6cos(2t),

y=t5/2.

Calculate the equation of the tangent to this curve at the point given by t=π/4, in the form y=mx+c.

The tangent is given by y=

Homework Equations

The Attempt at a Solution

[/B]
the answer that I got was y=-0.145x+0.547 but it says that it is wrong

I did this by subbing in the value of t for x and y and also the differential of x and y and then calculating the slope

but I am not sure what I have done wrong

 

[ehild]

Homework Statement

Consider the parametric curve given by:

x=6cos(2t),

y=t5/2.

Calculate the equation of the tangent to this curve at the point given by t=π/4, in the form y=mx+c.

The tangent is given by y=

Homework Equations

The Attempt at a Solution

[/B]
the answer that I got was y=-0.145x+0.547 but it says that it is wrong

I did this by subbing in the value of t for x and y and also the differential of x and y and then calculating the slope

but I am not sure what I have done wrong

How can we know what you did wrong if you do not show your work?

 

[Ray Vickson]

Homework Statement

Consider the parametric curve given by:

x=6cos(2t),

y=t5/2.

Calculate the equation of the tangent to this curve at the point given by t=π/4, in the form y=mx+c.

The tangent is given by y=

Homework Equations

The Attempt at a Solution

[/B]
the answer that I got was y=-0.145x+0.547 but it says that it is wrong

I did this by subbing in the value of t for x and y and also the differential of x and y and then calculating the slope

but I am not sure what I have done wrong

Do you mean $y = (5/2)t$, $y = t^{5/2}$, $y = t^5/2 = \frac{1}{2} t^5$, or what? Use parenthesis, like this: y = (5/2)t or y = t*(5/2) or y = t^(5/2) or y =(t^5)/2, or whatever is appropriate.

 

[53Mark53]

Do you mean $y = (5/2)t$, $y = t^{5/2}$, $y = t^5/2 = \frac{1}{2} t^5$, or what? Use parenthesis, like this: y = (5/2)t or y = t*(5/2) or y = t^(5/2) or y =(t^5)/2, or whatever is appropriate.

sorry the question didnt copy properly it should say t^(5/2)