Calculating the Tangent to a Parametric Curve at a Given Point

In summary, the problem involves finding the equation of the tangent to a parametric curve given by x=6cos(2t) and y=t^(5/2), at the point t=π/4. The solution involves subbing in the value of t for x and y, taking the differential of x and y, and calculating the slope. However, the answer provided (y=-0.145x+0.547) was deemed incorrect without showing the work. Clarification is needed on the specific form of y in the problem.
  • #1
53Mark53
52
0

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
 
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  • #2
53Mark53 said:

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
53Mark53 said:

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
Ray Vickson said:
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)
 

1. What is a parametric curve?

A parametric curve is a mathematical function that describes the relationship between two variables, usually represented by the letters x and y. The curve is defined by a set of parametric equations, which express the x and y coordinates of points on the curve in terms of a third variable, typically represented by the letter t.

2. How is the tangent of a parametric curve calculated?

The tangent of a parametric curve at a given point can be calculated by finding the derivatives of the parametric equations with respect to the parameter t and then evaluating those derivatives at the desired point. The slope of the tangent line is equal to the ratio of these derivatives (dy/dt divided by dx/dt).

3. Why is the tangent important in parametric curves?

The tangent of a parametric curve represents the instantaneous rate of change at a given point on the curve. This is important because it allows us to determine the direction and steepness of the curve at that point. It also has practical applications, such as in engineering where it can help determine the velocity and acceleration of moving objects along a parametric path.

4. How do I find the equation of the tangent line to a parametric curve?

To find the equation of the tangent line at a specific point on a parametric curve, you can follow these steps: 1) Find the derivatives of the parametric equations with respect to the parameter t. 2) Plug in the t-value of the desired point to find the slope of the tangent line. 3) Use the point-slope formula (y - y1 = m(x - x1)) to find the equation of the tangent line, where (x1, y1) is the point of tangency and m is the slope.

5. Can the tangent of a parametric curve be negative?

Yes, the tangent of a parametric curve can be negative. This means that the curve is decreasing at that point, with a slope that is moving downward. However, the tangent can also be positive, indicating a curve that is increasing at that point with an upward slope. It is important to note that the tangent value does not necessarily determine the direction of the curve overall, as it can change throughout the curve.

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