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Parametric curves

  • Thread starter mjstyle
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  • #1
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Homework Statement


a)Consider the parametric curve x = t^2 + t, y = e^t. Find all t such that the tangent line of the curve at (x(t), y(t)) intersects the x-axis at (4,0)

Homework Equations





The Attempt at a Solution


I draw out the graph and came out with the points, I was wondering do i eliminate the parameter t to find the equation first or what approach can i take to start this question. Thank you
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi mjstyle ! Welcome to PF! :smile:
a)Consider the parametric curve x = t^2 + t, y = e^t. Find all t such that the tangent line of the curve at (x(t), y(t)) intersects the x-axis at (4,0)

I draw out the graph and came out with the points, I was wondering do i eliminate the parameter t to find the equation first or what approach can i take to start this question. Thank you
The parameter t is the answer, so eliminate it last. :wink:

Start by writing down the slope of the line from (4,0) to (x(t),y(t)), then get it to equal the slope of the tangent at (x(t),y(t)) …

what do you get? :smile:
 
  • #3
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Thank you so much for the quick response,

I was wondering, writing down the slope of the line from (4,0) to (x(t),y(t)),

m = y(t) - 0 / x(t) - 4

tangent at (x(t),y(t)):

dy/dx = dy/dt / dx/dt = e^t / 2t + 1

then i'm stuck hehe
 
  • #4
tiny-tim
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m = y(t) - 0 / x(t) - 4

then i'm stuck hehe
hmm :rolleyes: … that's because (unlike your dy/dx) you haven't yet converted m into a function of just t :wink:
 
  • #5
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omg what am i thinking, that's right

m = y(t) - 0 / x(t) - 4

m = e^t / t^2 + t - 4

tangent at (x(t),y(t)):

dy/dx = dy/dt / dx/dt = e^t / 2t + 1

x = t^2 + t, y = e^t

so right right i fond the slop of the line according to t, how do i find the slop of the tangent line: x = t^2 + t, y = e^t?

thanks
 
  • #6
tiny-tim
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… how do i find the slop of the tangent line …
the slope of the tangent line is dy/dx :smile:
 
  • #7
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oh... i always thought that's the tangent line dy/dx then what i do now is

e^t / 2t + 1 = e^t / t^2 + t - 4 and solve for t?

which is

ln t / 2t + 1 = ln t / t^2 + t - 4

t^2 + t - 4 / 2t + 1 = 1

t^2 - t - 5 = 0

using quadratic formula: comes out to be (1 + square root(21)) / 2 and (1 - square root(21)) / 2

is that correct? thank you so much
 
  • #8
tiny-tim
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(try using the X2 tag just above the Reply box :wink:)

Yes, but it would have been a lot quicker just to divide your original equation by et

(and your "ln t / 2t + 1 = ln t / t^2 + t - 4" is rubbish :rolleyes:)
 
  • #9
8
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thank you so much for the help!. I actually have one more question,

Determine the values of t for which the curve x = 2squareroot(1+t), y = intergral from x to t^2 (squareroot(u) - 1)squareroot(1 + squareroot(u)) du, t greater and equal to 0 is concave upward and those for thich is it concave downward.

Do i find the second derivative of this and then.... hehe
 
  • #10
tiny-tim
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… Do i find the second derivative …
(have a square-root: √ and an integral: ∫ :wink:)

Yes, you need the sign (only) of d2y/dx2

but since x is monotone increasing (wrt t), that'll be the same as the sign of d/dt (dy/dx) :smile:
 
  • #11
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hey tiny-tim, for the preivous question, i don't think it's right cause i plugged in 1+squareroot(21) / 2 into t in y = e^t and i did not get 0....
 
  • #12
tiny-tim
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(what happened to that √ i gave you?)
hey tiny-tim, for the preivous question, i don't think it's right cause i plugged in 1+squareroot(21) / 2 into t in y = e^t and i did not get 0....
sorry … not following you :redface:
 
  • #13
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oh cause, the question asks

Find all t such that the tangent line of the curve at (x(t), y(t)) intersects the x-axis at (4,0)

so i'm assuming the t's I fins intersects the x-axis, so let say i plug in the t i find into the y(t) equation, shouldn't i get 0, but plugging in 1+squareroot(21) / 2 into t in y = e^t doesn't give me 0.
 
  • #14
tiny-tim
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let say i plug in the t i find into the y(t) equation
?? :confused:

if you plug in a t you found into the y(t) equation, you get the y coordinate of a point on the curve from which the tangent goes through (4,0)
 

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