Differentiating Parametric Equations: x = t/2, y = t^2 + 1

[warwick]

I don't know why I'm not getting the second derivative questions right.

1. a) Find the slope of the tangent line to the parametric curve x = t/2, y = t^2 + 1 at t = -1 and t = 1 without eliminating the parameter.
b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

dy/dx = 2t/(1/2) = 4t

y'(1) = 4
y'(-1) = -4

x = t/2, y - 1= t^2

(2x)^2 = t^2

y - 1 = 4x^2

y = 4x^2 + 1



5 - 10. Find dy/dx and d^2y/dx^2 at the given point without eliminating the parameter.

5. x = t^1/2, y = 2t + 4; t = 1

dy/dx = 2/(1/2)t^(-1/2) = 4t^1/2

y'(1) = 4

y'' = 4

y''(1) = 2

7. x = sec t, y = tan t; t = pi/3

dy/dx = csc x

y'(pi/3) = 2/3^1/2

y''= -1/(3 square root of 3)

9. x = theta + cos theta, y = 1 + sin theta; theta = pi/6

dy/dx = cos theta/1 - sin theta

y'(pi/6) = 3^1/2

y'' =

 

[Deleted member 4993]

Re: Differentiating Parametric Equations: x = t/2, y = t^2 +

5 - 10. Find dy/dx and d^2y/dx^2 at the given point without eliminating the parameter.

5. x = t^1/2, y = 2t + 4; t = 1

dy/dx = 2/(1/2)t^(-1/2) = 4t^1/2

y" = d/dx [dy/dx] = {d/dt[dy/dx]}/{dx/dt} = {4 * 1/2 * t^(-1/2)}/{1/2 * t^(-1/2)} = 4

 

[warwick]

Ok. Let me try the rest of them.

What about number 1? I'm getting incorrect derivative with respect to x.

 

[warwick]

9. x = theta + cos theta, y = 1 + sin theta; theta = pi/6

dy/dx = cos theta/1 - sin theta

y'(pi/6) = 3^1/2

y'' = [- sin theta + 1]/[(1 - sin theta)^2 (1 - sin theta)]

y''(pi/6) = 4

 

[Deleted member 4993]

9. x = theta + cos theta, y = 1 + sin theta; theta = pi/6

dy/dx = cos theta/1 - sin theta

y'(pi/6) = 3^1/2

y'' = [- sin theta + 1]/[(1 - sin theta)^2 (1 - sin theta)] <--- Cancel out common factors to simplify
y''(pi/6) = 4

 

[warwick]

9. x = theta + cos theta, y = 1 + sin theta; theta = pi/6

dy/dx = cos theta/1 - sin theta

y'(pi/6) = 3^1/2

y'' = [- sin theta + 1]/[(1 - sin theta)^2 (1 - sin theta)] <--- show your work step by step - I do not get the same answer
y''(pi/6) = 4

I found my error and corrected it.

 

[Deleted member 4993]

Ok. Let me try the rest of them.

What about number 1? I'm getting incorrect derivative with respect to x.

Show detailed work...

 

[warwick]

Ok. Let me try the rest of them.

What about number 1? I'm getting incorrect derivative with respect to x.

Show detailed work...

I did in my original post. Once again, I'm probably making a careless computational error.

 

[Deleted member 4993]

Ok. Let me try the rest of them.

What about number 1? I'm getting incorrect derivative with respect to x.

Show detailed work...

I did in my original post. Once again, I'm probably making a careless computational error.

I don't see that you have completed the task - and where is the problem?

 

[warwick]

[quote="Subhotosh Khan":yu2oildy]

Ok. Let me try the rest of them.

What about number 1? I'm getting incorrect derivative with respect to x.

Show detailed work...

I did in my original post. Once again, I'm probably making a careless computational error.

I don't see that you have completed the task - and where is the problem?[/quote:yu2oildy]

1. a) Find the slope of the tangent line to the parametric curve x = t/2, y = t^2 + 1 at t = -1 and t = 1 without eliminating the parameter.
b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

dy/dx = 2t/(1/2) = 4t

y'(1) = 4
y'(-1) = -4

x = t/2, y - 1= t^2

(2x)^2 = t^2

y - 1 = 4x^2

y = 4x^2 + 1

 

[Deleted member 4993]

I don't see that you have completed the task - and where is the problem?

1. a) Find the slope of the tangent line to the parametric curve x = t/2, y = t^2 + 1 at t = -1 and t = 1 without eliminating the parameter.
b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

dy/dx = 2t/(1/2) = 4t

y'(1) = 4
y'(-1) = -4

x = t/2, y - 1= t^2

(2x)^2 = t^2

y - 1 = 4x^2

y = 4x^2 + 1

Finish your checking by putting in appropriate values of variables.

 

[warwick]

[quote="Subhotosh Khan":1tjfbi19]

I don't see that you have completed the task - and where is the problem?

1. a) Find the slope of the tangent line to the parametric curve x = t/2, y = t^2 + 1 at t = -1 and t = 1 without eliminating the parameter.
b) Check your answers in part (a) by eliminating the parameter and differentiating an appropriate function of x.

dy/dx = 2t/(1/2) = 4t

y'(1) = 4
y'(-1) = -4

x = t/2, y - 1= t^2

(2x)^2 = t^2

y - 1 = 4x^2

y = 4x^2 + 1

Finish your checking by putting in appropriate values of variables.

[/quote:1tjfbi19]

dy/dx = 8x

x(1) = 1/2
x(-1) = -1/2

y'(1) = 4
y'(-1) = -4

 

[warwick]

11. (a) Find the equation of the tangent line to the curve x = e^t, y = e^(-t) at t = 1 without eliminating the parameter.

dy/dx = -e^t/e^t = -1

y - e^(-1) = - (x - e)

y = - x + e + e^(-1)

(b) Find the equation of the tangent line in part (a) b eliminating the parameter.

ln x = t, ln y = - t

ln x = ln y

23. Find the slope of the tangent line to the polar curve for the given value of theta.

r = 1/theta; theta = 2

dr/dtheta = - 1/theta^2

dy/dx = [(1/2) cos 2 + sin 2 (- theta^-2)]/[ (-1/2) sin 2 + cos 2 (- theta^-2)]