Question

Find an equation of the tangent to the curve at the given point by two methods: (a) without eliminating the parameter and b) by first elminating the parameter

Answer 1

\[x = 1 + lnt, y = t^2 + 2\]

Answer 2

At (1, 3)

Answer 3

For part a,\[\large\rm \frac{dy}{dx}=\frac{\dot y}{\dot x}\]

Answer 4

I know how to do this problem given a value for t, but not given the point...

Answer 5

For dy/dx I got 2t^2

Answer 6

What are the dots on top of the y and x?

Answer 7

It's just a different notation used for `time derivatives`.
\(\large\rm \frac{dy}{dt}=\dot y=y'\)

Answer 8

Ah interesting. Okay

Answer 9

\[\large\rm \color{orangered}{x}=1+\ln t\qquad\qquad\qquad \color{orangered}{y}=t^2+2\]We have values for x and y at least,\[\large\rm \color{orangered}{1}=1+\ln t\qquad\qquad\qquad \color{orangered}{3}=t^2+2\]

Answer 10

This should allow you to find values for t.

Answer 11

ln(t)=0 has no solution.
Are you sure you have the equations written down properly? Hmm

Answer 12

Hmm yes the equations are correct. The book gives the answer as: y = 2x + 1

Answer 13

Did I say ln(t)=0 has no solution?
Oh boy my brain is failing today :) lol.

I meant ln(0) has no solution.
But obviously ln(t)=0 has the solution of t=1, right?

Answer 14

You have to make sure that value satisfies both equations though.

Answer 15

Your derivative looks correct.\[\large\rm \frac{dy}{dx}=2t^2\]

Answer 16

Evaluating this derivative at x=1, y=3 and t=1,\[\large\rm \frac{dy}{dx}=2\]This gives us the `slope` of the tangent line, ya?\[\large\rm y-y_o=m(x-x_o)\]\[\large\rm y-y_o=2(x-x_o)\]

Answer 17

Where did you get the values x=1, y=3 and t=1 from?

Answer 18

(x,y) = (1,3) corresponds to some `specific t value`.

Answer 19

x and y take on these values at some `specific time`.

Answer 20

By plugging x=1 and y=3 into our equations,
we are able to solve for t.

Answer 21

\[\large\rm \color{orangered}{x}=1+\ln t\qquad\qquad\qquad \color{orangered}{y}=t^2+2\]\[\large\rm \color{orangered}{1}=1+\ln t\qquad\qquad\qquad \color{orangered}{3}=t^2+2\]

Answer 22

\[\large\rm 3=t^2+2\]Subtracting 2,\[\large\rm 1=t^2\]Square root,\[\large\rm t=\pm1\]So t could potentially be 1 or negative.
We need to find the value that satisfies BOTH equations though.\[\large\rm 1=1+\ln t\]\[\large\rm 0=\ln t\]t=-1 will NOT satisfy this equation.
So t=1 is the specific time that we were looking for.

Answer 23

Gotcha! Now all that's left to do is find the slope at t = 1, which is 2.
y - 3 = 2(x-1)
y = 2x +1
Yep :) thanks zep!

Answer 24

What about for part b though?

Answer 25

\[\large\rm x=1+\ln t\qquad\qquad\qquad y=t^2+2\]Hmm..
Let's try solving for t in the first equation.\[\large\rm x-1=\ln t\]\[\large\rm e^{x-1}=t\]And I guess we can replace the t in our second equation with this, ya?

Answer 26

Yes that sounds right

Answer 27

\[y = e ^{2x-2}+2\]

Answer 28

Like this right?

Answer 29

Umm yes.
So doing it this way, you've completely eliminated the parameter.
No need for your fancy parameter derivative.. thing.

Answer 30

So would I take the derivative the regular way?

Answer 31

Yes

Answer 32

Does this look right?
\[(4x-4)e ^{2x-2}\]

Answer 33

\[\large\rm y'=e^{2x-2}(2x-2)'\]

Answer 34

Wouldn't it need to be multiplied by the inside function? So by 2?

Answer 35

*The derivative of the inside function I mean

Answer 36

\[\large\rm y'=e^{2x-2}(2)\]Ya that looks better.

Answer 37

... but the top still needs to come down right? Like
\[y' = (2x-2)e ^{2x-2}*2\]

Answer 38

Or am I wrong?

Answer 39

No, the top doesn't come down.
The derivative of the top comes down only.

Answer 40

Gotta master that chain rule girly!! :)

Answer 41

Mmmmm yeah, the e^x and e^a stuff always trips me up.
So \[2e ^{2x-2}\]
Then what?

Answer 42

\(\large\rm y'(x)=2e^{2x-2}\)
Then evaluate your derivative function at x=1, y=3.

Answer 43

By plugging in x =1?

Answer 44

It would equal 2. Then the same steps as last time to get the equation. Alrighty :) Zepman to the rescue.

Answer 45

yayyy team

Answer 46

:D