Why doesn't this work

[Loki123]

There are two curves. I have written them at the top of the first picture. The question is at what angle do they cross. I got a correct answer, however, in the first picture I plug in intercepts before the derivative and get 0 degrees. But when I plug them after I get derivative (second picture), I get the correct answer 90 degrees. Why can't I plug them in before the derivative??

 

[Dr.Peterson]

There are two curves. I have written them at the top of the first picture. The question is at what angle do they cross. I got a correct answer, however, in the first picture I plug in intercepts before the derivative and get 0 degrees. But when I plug them after I get derivative (second picture), I get the correct answer 90 degrees. Why can't I plug them in before the derivative??
View attachment 32393View attachment 32394

First, why are you finding x- and y-intercepts at all? How is that related to the question?

Second, when you find that (for the first curve) [imath]a=-1[/imath] or [imath]x=-\frac{1}{2}[/imath], these are two different ways in which the equation can be true with y=0. The first tells you that in the special case [imath]a=-1[/imath], the graph is the horizontal line [imath]y=0[/imath], not just a single point; x can have any value. It looks like this (the red line):

​

For any other value of the parameter a, the x-intercept is at -1/2, as in this example with [imath]a=-0.9[/imath]:

​

But, again, the intercepts are not part of the problem. The point that matters is the intersection of the two curves, when [imath]x=-\frac{a}{2}[/imath].

And in that case, you need to keep in mind that [imath]a[/imath] is a parameter, which is to be thought of as a fixed number. The intersection is not a value of a, but a value of x. Don't express it as [imath]a=-2x[/imath]. So all that you do on the bottom of the first page is irrelevant.

 

[Steven G]

Why are you multiplying out?
(a+1)(2x+1) = 0 if (a+1)=0 or 2x+1=0. Then a=-1 or x=-1/2.

 

[Steven G]

y^2 = (a+1)(2x+1)
2yy' = 2(a+1)
y' = (a+1)/y = ....

This is easier than solving for y and dealing with square roots while computing the derivative.

 

[Loki123]

First, why are you finding x- and y-intercepts at all? How is that related to the question?

Second, when you find that (for the first curve) [imath]a=-1[/imath] or [imath]x=-\frac{1}{2}[/imath], these are two different ways in which the equation can be true with y=0. The first tells you that in the special case [imath]a=-1[/imath], the graph is the horizontal line [imath]y=0[/imath], not just a single point; x can have any value. It looks like this (the red line):

View attachment 32395​

For any other value of the parameter a, the x-intercept is at -1/2, as in this example with [imath]a=-0.9[/imath]:

View attachment 32396​

But, again, the intercepts are not part of the problem. The point that matters is the intersection of the two curves, when [imath]x=-\frac{a}{2}[/imath].

And in that case, you need to keep in mind that [imath]a[/imath] is a parameter, which is to be thought of as a fixed number. The intersection is not a value of a, but a value of x. Don't express it as [imath]a=-2x[/imath]. So all that you do on the bottom of the first page is irrelevant.

i mirrored this, that's why i did it the way I did.

Angle two curves

At what angle do curves y^2=4a(a-x) and y^2=4b(b+x) cross each other. a and b are real numbers. I tried to get coefficients and then calculate the angle but It didn't work.

www.freemathhelp.com

 

[BigBeachBanana]

i mirrored this, that's why i did it the way I did.

Angle two curves

At what angle do curves y^2=4a(a-x) and y^2=4b(b+x) cross each other. a and b are real numbers. I tried to get coefficients and then calculate the angle but It didn't work.

www.freemathhelp.com

You're correct that the two curves intercept at [imath]x=-a/2[/imath]. Now I don't understand how you got the y values. To find the y values, simply plug the x-value into one of the original equations.
[math]y^2=(a+1)(2x+1)=(a+1)(2\left(-\frac{a}{2}\right)+1)=(1-a^2)\\ \implies y=\pm\sqrt{1-a^2}\\ \therefore \text{intercepts are } (-\frac{a}{2},\pm\sqrt{1-a^2})[/math]

 

[Loki123]

You're correct that the two curves intercept at [imath]x=-a/2[/imath]. Now I don't understand how you got the y values. To find the y values, simply plug the x-value into one of the original equations.
[math]y^2=(a+1)(2x+1)=(a+1)(2\left(-\frac{a}{2}\right)+1)=(1-a^2)\\ \implies y=\pm\sqrt{1-a^2}\\ \therefore \text{intercepts are } (-\frac{a}{2},\pm\sqrt{1-a^2})[/math]

right, i forgot to find y intercept, the y values you see are for x=0

 

[BigBeachBanana]

right, i forgot to find y intercept, the y values you see are for x=0

As Dr.P mentioned, there's no need for the y-values at x=0. Again, I'll use implicit differentiation for the derivative since it's a lot cleaner. I recommend you learn this because it's handy for problems like this. This should be something you learned in differential calculus.
[math]y_1^2=(a+1)(2x+1)\\ 2y\cdot y_1'=(a+1)(2)\\ y_1'=\frac{a+1}{y}\\[/math]Similarly,
[math]y_2^2=(a-1)(2x-1)\\ 2y\cdot y_2'=(a-1)(2)\\ y_2'=\frac{a-1}{y}[/math]First point of intercept [imath]\left(-\frac{a}{2},\sqrt{1-a^2}\right)[/imath]
[math]y_1'=\frac{a+1}{\sqrt{1-a^2}}[/math][math]y_2'=\frac{a-1}{\sqrt{1-a^2}}[/math][math]\tan(\theta_1)=\frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a-1}{\sqrt{1-a^2}}}{1+\frac{a+1}{\sqrt{1-a^2}}\cdot \frac{a-1}{\sqrt{1-a^2}}}=0\\ \implies \theta_1=\pi=180\degree [/math]Repeat the process for the second intercept.

 

[BigBeachBanana]

[math]\cancel{\tan(\theta_1)=\frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a-1}{\sqrt{1-a^2}}}{1+\frac{a+1}{\sqrt{1-a^2}}\cdot \frac{a-1}{\sqrt{1-a^2}}}=0}\\ [/math]

I made a boo boo at the last step.
[math]\tan(\theta_1)=\frac{\text{[stuff]}}{0}=\text{undefined}\\ \implies \theta_1=\frac{\pi}{2}=90\degree[/math]

 

[topsquark]

I made a boo boo at the last step.
[math]\tan(\theta_1)=\frac{\text{[stuff]}}{0}=\text{undefined}\\ \implies \theta_1=\frac{\pi}{2}=90\degree[/math]

The corner is starting to get crowded.

-Dan

 

[BigBeachBanana]

The corner is starting to get crowded.

-Dan

To address the overcrowding situation, I petition to expand to the basement. However, it's inhabitable at the moment. Therefore, I shall spend my time rehabilitating the basement, in lieu of staying in the corner, while awaiting for future members to come and join me.

PS: It's hot down here.

 

[Deleted member 4993]

To address the overcrowding situation, I petition to expand to the basement. However, it's inhabitable at the moment. Therefore, I shall spend my time rehabilitating the basement, in lieu of staying in the corner, while awaiting for future members to come and join me.

PS: It's hot down here.

But I thought you were at the beach......