# Solving Parametric Curve: Find t for x=4, y=0

• mjstyle
In summary, the parametric curve x = 2squareroot(1+t), y = e^t has a slope of dy/dx = e^t / 2t + 1 and a tangent line at (x(t), y(t)) that slopes dy/dx = e^t / 2t + 1. The t such that the tangent line at (x(t), y(t)) intersects the x-axis at (4,0) is t^2 + t.f

## 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)

## 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

Welcome to PF!

Hi mjstyle ! Welcome to PF!
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.

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?

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

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

then I'm stuck hehe

hmm … that's because (unlike your dy/dx) you haven't yet converted m into a function of just t

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

… how do i find the slop of the tangent line …

the slope of the tangent line is dy/dx

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

(try using the X2 tag just above the Reply box )

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 )

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

… Do i find the second derivative …

(have a square-root: √ and an integral: ∫ )

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)

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...

(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