Question

If the curve represented by x=sec^2 t and y=cost and tangent at P,where t=pi/4 intersect the curve again at Q,then find PQ.

Answer 1

@hartnn

Answer 2

So, you're given the two parametric equations
\[\begin{cases}x=\sec^2t\\y=\cos t\end{cases}\]
And you're also told that the curve at point P (determined by t=\(\dfrac{\pi}{4}\)) has a tangent line that intersects the curve again at Q. Is that right?

And the question asks you to find PQ: is that supposed to be a numerical answer (as in the distance between P and Q), or the equation of the line containing the segment PQ (Cartesian or parametric ?) ?

Answer 3

I'd say the first thing to do is find P.
\[\begin{cases}x=\sec^2t\\y=\cos t\end{cases}~\Rightarrow~P=\left(\sec^2\frac{\pi}{4},\cos\frac{\pi}{4}\right)=\left(\frac{1}{2},\frac{1}{\sqrt2}\right)\]
Next, find the slope of the tangent line when \(t=\dfrac{\pi}{4}\):
\[\frac{dy}{dx}=\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }~\Rightarrow~\frac{dy}{dx}_{~~t=\pi/4}=\frac{ -\sin t }{ 2\sec t \cdot\sec t\tan t }_{~~t=\pi/4}=\cdots\]

Answer 4

dudee u messed up!!

Answer 5

If the curve represented by x=sec^2 t and y=cost and tangent at P,where t=pi/4 intersect the curve again at Q,then find PQ.
$$x=\sec^2 t,y=\cos t\\\frac{dx}{dt}=2\sec t\sec t\tan t=2\sec^2\tan t,\frac{dy}{dt}=-\sin t\\\frac{dy}{dx}=\frac{-\sin t}{2\sec^2 t\tan t}=-\frac{\cos t}{2\sec^2 t}=-\frac12\cos^3 t$$

Answer 6

so at \(t=\frac{\pi}4\) we have:$$\frac{dy}{dx}=-\frac12\left(\frac{\sqrt2}2\right)^3=-\frac{2\sqrt2}{16}=-\frac{\sqrt2}8$$

Answer 7

and \(x=\sec^2 \pi/4=2,y=\cos\pi/4=\dfrac{\sqrt2}2\)

Answer 8

so we can parameterize this tangent line as follows:$$x=2-\frac{\sqrt2}8t\\y=\frac{\sqrt2}{2}-\frac{\sqrt2}8t$$

Answer 9

:O well that is complex

Answer 10

oops i forgot i'm using two different parameterizations...

Answer 11

I suppose it's best if we convert to rectangular now... our tangent line becomes$$y-\frac{\sqrt2}2=-\frac{\sqrt2}8(x-2)$$and our original curve:$$y=\cos x=\frac1{\sec x}=\pm\frac1{\sqrt{\sec^2 x}}=\pm\frac1{\sqrt{x}}$$

Answer 12

h$$\frac{\sqrt2}2-\frac{\sqrt2}8(x-2)=\pm\frac1{\sqrt x}\\\frac1{32}x^2-\frac38x+\frac98=\frac1x\\x^2-12x+36=\frac1x\\x^3-12x^2+36x-1=0$$

Answer 13

OOPS$$x^3-12x^2+36x=32\\x^3-12x^2+36x-32=0$$

Answer 14

now, we know our tangent line touches at \(x=2\) hence we expect that to be a solution to our above equation... sure enough:$$8-12(4)+36(2)-32=8-12(2)-32=8-8=0$$

Answer 15

now we can divide out \(x-2\) to yield a quadratic$$\begin{align*}x^3-12x^2+36x-32&=x(x^2-12x+36)-32\\&=(x-2)(x^2-12x+36)+2(x^2-12x+36)-32\\&=(x-2)(x^2-12x+36)+2(x^2-12x+36-16)\\&=(x-2)(x^2-12x+36)+2(x^2-12x+20)\\&=(x-2)(x^2-12x+36)+2(x-2)(x-10)\\&=(x-2)(x^2-12x+36+2(x-10))\end{align*}$$hence dividing by \(x-2\) we're left with $$x^2-12x+36+2(x-10)=x^2-10x+16=(x-8)(x-2)=0$$so ignoring the repeated root \(x=2\) observe that our second point of intersection is just \(x=8\)

Answer 16

To determine our \(y\), we plug in to either the equation of our curve or our tangent line:$$\frac{\sqrt2}2-\frac{\sqrt2}8(8-2)=\frac{\sqrt2}2-\sqrt2+\frac{\sqrt2}4=-\frac{\sqrt2}{4}$$

Answer 17

... now our distance is just:$$\sqrt{(8-2)^2+\left(\frac{\sqrt2}2+\frac{\sqrt2}4\right)^2}=\sqrt{6^2+\left(\frac{3\sqrt2}4\right)^2}=\sqrt{36+\frac98}=\frac34\sqrt{66}\approx6.09$$

Answer 18

@DLS does that agree with your answer?

Answer 19

yup

Answer 20

yw

Answer 21

I didn't say thanks lol

Answer 22

well you ought to considering I just did the problem

Answer 23

hmm :P