1

The question asks to find the equation of the tangent to this curve at the point $t=\pi/4$.

I've determined $$\frac{dy}{dx} =(\frac{dy}{dt})/(\frac{dx}{dt}) = -0.222$$ Have I got the right idea?

Also asks for the solution to be in the form $y=mx+c$, thank you.

Rellek
  • 2,222
AidanLeith
  • 21
  • yes you are correct so far. So assuming that you are looking to also solve for the line, you can use your previous knowledge of how to obtain the equation given the gradient $m$ and a point i.e. $(x,y)$ at which the curve must intersect. – Chinny84 May 18 '15 at 12:50
  • as far as I understand, m=gradient=-0.222, and c=y-intercept, but how do I figure out the y-intercept in this case? – AidanLeith May 18 '15 at 12:52
  • there is an answer down below. But essentially you have $x(\pi/4),y(\pi/4)$ and with $m$ leaves only one unknown. – Chinny84 May 18 '15 at 12:54
  • using that method I get c=(pi/4)^(3/2), therefore y=-0.222x+(pi/4)^(3/2), but it's saying the solution is incorrect, can you confirm that the answer's are correct? or have I made a mistake – AidanLeith May 18 '15 at 13:12
  • Indeed, I don't get what you got also. what did you get for $x$ and $y$? also did you confirm that the gradient was correct? – Chinny84 May 18 '15 at 13:22
  • It would be great if you also told us on which interval this curve is defined. – Alex M. May 18 '15 at 13:41
  • 1
    @AidanLeith general comment: check this reference guide for how to format math on this site--as is it's sort of unreadable. – MichaelChirico May 18 '15 at 13:41
  • 1
    @Aidan: You might add the details of how you computed the slope of the tangent line, since the slope should come out to $-\sqrt{\pi/8} \approx -0.6266...$. – Andrew D. Hwang May 18 '15 at 14:12

3 Answers3

1

first find the $x$ and $y$ coordinates of the point at time $t = \pi/4.$ we have $x(\pi/4) = 3\sqrt 2/2 = 2.121, y = \left(\frac{\pi}4\right)^{3/2} = 0.696.$

now evaluate the slope at this point $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}\big|_{t = \pi/4} = \frac{-3\sin \pi/4}{3\sqrt 2/2} = -1$$ therefore the tangent line at $t = \pi/4$ is $$y-0.696 = -1(x -2.121)\to y = 2.817-x $$

abel
  • 29,170
0

The direction vector of the tangent you're looking for is given by $(\frac{\mathbb d x}{\mathbb d t},\frac{\mathbb d y}{\mathbb d t})$, evaluated in $\frac \pi 4$. Then you can determine $m$. Finally you deduce $c$ by taking $t = \frac \pi 4$ : since $(x(\frac \pi 4),y(\frac \pi 4))$ is a point of the tangent, $y(\frac \pi 4) = m x(\frac \pi 4) + c$.

Alex M.
  • 35,207
Augustin
  • 8,446
  • using this solution I get c=(pi/4)^(3/2), but inputting this solution (y=-0.222x+(pi/4)^(3/2), says that the solution is wrong, can you confirm I've done the question correctly? – AidanLeith May 18 '15 at 13:07
0

Hints: The usual way of finding a tangent line to a parametric curve $(x(t), y(t))$ at some number $t_{0}$ is to compute

  • A point on the line, such as the point of tangency $(x_{0}, y_{0}) = (x(t_{0}), y(t_{0}))$.

  • The slope of the tangent line, which is $m = y'(t_{0})/x'(t_{0})$ if $x'(t_{0}) \neq 0$.

Then use the point-slope form $$ y - y_{0} = m(x - x_{0}),\quad\text{or}\quad y = mx + (y_{0} - mx_{0}). $$

Andrew D. Hwang
  • 78,195