Parametric curve and some irregular integtrals

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The discussion focuses on solving problems related to a parametric curve defined by the equations x = t^2 + 3t and y = 4 - t^2. For part 1a, the tangent line at the point (0, -5) is derived with a slope of -6/5, leading to the equation y = -6/5(x) - 5. In part 1b, vertical tangents are found at t = -2 and t = 2, corresponding to points (-2, 0) and (10, 0). The discussion also addresses the convergence of two definite integrals, with the first integral diverging and the second requiring substitution for evaluation.

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Homework Statement



1) Consider the paramerric curve given by x = t^2 + 3t and y = 4 - t^2
a) Find an equation of the tangent lne to the curve at the point (x,y) = (0,-5)
b) Determine the equation of every vertical tangent line to this parametric curve.

2)For each of the following definite integrals, determine its value if it convereges, otherwise, explain why it diverges.

a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]
b) integral (-2 to -3) [dx/(sqrt(x^2 - 4))]

The Attempt at a Solution



1)

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

since x = t^2 + 3t and y = 4 - t^2 and (x,y) = (0,-5)

t = -3

dy/dx = (-2*-3)/(4 - (-3)^2) = -(6/5)

y = -6/5(x) -5

b) 4 - t^2 = 0

t = -2, 2

(x,y) = (-2, 0) and (10, 0)

I can figure out the equation if these points are correct

2)

a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]

I get

-2 lim (t -> inf) [ 1 / ln(x)^1/2 ] ( e - t)



b) integral (-3 to -2) [dx/(sqrt(x^2 - 4))]

I got as far as lim ( t -> -2) [ln | x + sqrt(x^2 - 4) | - ln 2 ] ( -3 to t)

any support is much appreciated.

Thanks
 
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sapiental said:

Homework Statement



1) Consider the paramerric curve given by x = t^2 + 3t and y = 4 - t^2
a) Find an equation of the tangent lne to the curve at the point (x,y) = (0,-5)
b) Determine the equation of every vertical tangent line to this parametric curve.

2)For each of the following definite integrals, determine its value if it convereges, otherwise, explain why it diverges.

a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]
b) integral (-2 to -3) [dx/(sqrt(x^2 - 4))]

The Attempt at a Solution



1)

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

since x = t^2 + 3t and y = 4 - t^2 and (x,y) = (0,-5)

t = -3

dy/dx = (-2*-3)/(4 - (-3)^2) = -(6/5)

y = -6/5(x) -5
How did you get dy/dx= -2t/(4-t^2)? That is
[tex]\frac{\frac{dy}{dt}}{y}[/tex]
Shouldn't x have something to do with dy/dx?

b) 4 - t^2 = 0

t = -2, 2

(x,y) = (-2, 0) and (10, 0)

I can figure out the equation if these points are correct
Those points are NOT correct because you have the denominator of the derivative wrong. What is the equation of ANY vertical line?

2)

a) integral (e to infinity) [dx/(x(ln(x))^(3/2))]

I get

-2 lim (t -> inf) [ 1 / ln(x)^1/2 ] ( e - t)
Again, how did you get that? The result of a definite integral (from e to t) with respect to x cannot contain an "x". And even if that were correct, you haven't yet answered the question! Does the integeral exist and if so what is it? Try the substitution u= ln(x)


b) integral (-3 to -2) [dx/(sqrt(x^2 - 4))]

I got as far as lim ( t -> -2) [ln | x + sqrt(x^2 - 4) | - ln 2 ] ( -3 to t)
Again, that is obviously incorrect- the result of the integral cannot contain "x"! What is the derivative of arcsin(x)?
 

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