Parametric curve tangent help

  • #1
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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
 

Answers and Replies

  • #2

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?
 
  • #3

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.
 
  • #4
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)
 

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