Tangent Line to f(x) Without Specified Point

[johnstobbart]

Homework Statement

Hello again.

The question asks me to find an equation of the tangent to the graph:

$$f(x)= - sin^2 x + 1/2, ~x~\epsilon~[0, \frac{\pi}{2}]$$

which makes an angle of 135° with the x-axis (measure anti-clockwise from the positive x-axis). Assume that the scales along the x- and y- axis are the same.

I just don't know how to find the point.

Homework Equations

The Attempt at a Solution

What I did was took the 135° value from the unit circle, which I think is:
$$\frac{3\pi}{4}$$
and worked with that as my x-coordinate.

The problem is that the question states that x ε [0, ∏/2], which means that 3∏/4 can't be x.

The only other value I can think of getting is:
$$\frac{\pi}{4}~=~\frac{\sqrt{2}}{2}$$ from $$sin(\pi - \frac{3\pi}{4})$$

What am I overlooking?

 

[JesseC]

I think you have misunderstood what the question is asking. It wants you to find the equation of a straight line, with a gradient given by the angle 135 degrees, which lies tangent to the curve. In a straight line of the form y = mx + b, what does an angle of 135 degrees mean the value of m is?

Immediately the word tangent makes be think: differentiation.

Differentiating the function will give you an equation of the gradient at every point along it. See if you can take it from there. You may need to use a trig formula.

 

[Simon Bridge]

Hint: What does the derivative of f(x) tell you?
[edit] Jesse beat me :(

 

[dydxforsn]

You will need: 2sin(x)cos(x) = sin(2x).

Also once you find your 'm' value for y = mx + b as JesseC mentioned you will then need to find your 'b' value. To do this you will need to go back and find out at exactly what point along the curve (what 'x' value) did the slope of the line make an angle of 135° with the x-axis (go back to your derivative equation and solve for 'x' with your slope value.) Plugging this value of 'x' into your original equation of $f(x) = - (\sin{x})^2 + \frac{1}{2}$ will give you the 'y' value at which this happened. You now have an (x, y) point that your tangent line must pass through. You've already found the slope, with the additional knowledge of what 'x' and 'y' have to be at a specific point it's a simple matter of solving for 'b'.

 

[johnstobbart]

I think you have misunderstood what the question is asking. It wants you to find the equation of a straight line, with a gradient given by the angle 135 degrees, which lies tangent to the curve. In a straight line of the form y = mx + b, what does an angle of 135 degrees mean the value of m is?

Immediately the word tangent makes be think: differentiation.

Differentiating the function will give you an equation of the gradient at every point along it. See if you can take it from there. You may need to use a trig formula.

I did misunderstand.

If that's the case, would the gradient be -√2/2?

 

[JesseC]

I did misunderstand.

If that's the case, would the gradient be -√2/2?

No, it is simpler than you are thinking. Perhaps it would help if you drew the problem. Do you have a protractor and a ruler handy? :D

Start by drawing a line that makes an angle of 135 degrees with the x axis.

Hint: m = 1 for angle = 45 degrees.

 

[Simon Bridge]

you need the relationship between a gradient and the angle.

the picture is a better way to show you ... but it is notoriously difficult to get people to draw pictures around here :(

 

[Ray Vickson]

Homework Statement

Hello again.

The question asks me to find an equation of the tangent to the graph:

$$f(x)= - sin^2 x + 1/2, ~x~\epsilon~[0, \frac{\pi}{2}]$$

which makes an angle of 135° with the x-axis (measure anti-clockwise from the positive x-axis). Assume that the scales along the x- and y- axis are the same.

I just don't know how to find the point.

Homework Equations

The Attempt at a Solution

What I did was took the 135° value from the unit circle, which I think is:
$$\frac{3\pi}{4}$$
and worked with that as my x-coordinate.

The problem is that the question states that x ε [0, ∏/2], which means that 3∏/4 can't be x.

The only other value I can think of getting is:
$$\frac{\pi}{4}~=~\frac{\sqrt{2}}{2}$$ from $$sin(\pi - \frac{3\pi}{4})$$

What am I overlooking?

As written, you have $f(x) = 1/2 - \sin^2 x.$ You did mean that, did you, rather than $f(x) = -\sin^2(x + 1/2)?$

RGV

 

[johnstobbart]

No, it is simpler than you are thinking. Perhaps it would help if you drew the problem. Do you have a protractor and a ruler handy? :D

Start by drawing a line that makes an angle of 135 degrees with the x axis.

Hint: m = 1 for angle = 45 degrees.

OK. I've drawn a line f(x) = -3x, which is 135 degrees to the x-axis using a graphing program.
I think I have it now. Now that I have the gradient, all I need to do is find the derivative of f(x), which is - 2sin x cos x. Then I make f '(x) = 3, and that should give my x value, right? Then how do I calculate sin 2x = -3/2?

As written, you have $f(x) = 1/2 - \sin^2 x.$ You did mean that, did you, rather than $f(x) = -\sin^2(x + 1/2)?$

RGV

Yes, I did mean $$\frac{1}{2} - sin^2 x$$

It is definitely not $$-sin^2(x + \frac{1}{2})$$

 

[JesseC]

OK. I've drawn a line f(x) = -3x, which is 135 degrees to the x-axis using a graphing program.

This is what happens when you don't use a pencil and paper and protractor :P. See the image I attached to this message.

It took me 30 seconds in paint. I have no idea how you got -3x on your graphing program.

I think I have it now. Now that I have the gradient, all I need to do is find the derivative of f(x), which is - 2sin x cos x. Then I make f '(x) = 3, and that should give my x value, right? Then how do I calculate sin 2x = -3/2?

This method will work if you use the correct value for m! You can tell that your value for m is wrong because sin(2x) must be equal to something in the range -1 to 1.

 

[johnstobbart]

This is what happens when you don't use a pencil and paper and protractor :P. See the image I attached to this message.
View attachment 49731
It took me 30 seconds in paint. I have no idea how you got -3x on your graphing program.

This method will work if you use the correct value for m! You can tell that your value for m is wrong because sin(2x) must be equal to something in the range -1 to 1.

Sorry for not using pencil and paper, but I thought a graph program would suffice. I got -3 because you said a gradient of 1 = 45°. 45 * 3 = 135°. I input y = 3x on the graph, which didn't seem to be 135°, so to get it in the next quadrant, I put in -3x to get a line that looked as though it was 135°.

Anyway, that's besides the point. After staring at your picture for a while, I think I clicked. Since a gradient of 1 is 45° to the x- axis, that means that -1 will have an angle of 135°. Solving for f '(x) = -1, I get x = ∏/4, which gives a y-value of 0. The equation of the tangent is y = -x + ∏/4.

I am hoping that's correct or I'm going to jump into a wall.

 

[dydxforsn]

The equation of the tangent is y = -x + ∏/4.

I am hoping that's correct or I'm going to jump into a wall.

It's right. Good job :)

 

[JesseC]

Sorry for not using pencil and paper, but I thought a graph program would suffice. I got -3 because you said a gradient of 1 = 45°. 45 * 3 = 135°. I input y = 3x on the graph, which didn't seem to be 135°, so to get it in the next quadrant, I put in -3x to get a line that looked as though it was 135°.

Anyway, that's besides the point. After staring at your picture for a while, I think I clicked. Since a gradient of 1 is 45° to the x- axis, that means that -1 will have an angle of 135°. Solving for f '(x) = -1, I get x = ∏/4, which gives a y-value of 0. The equation of the tangent is y = -x + ∏/4.

I am hoping that's correct or I'm going to jump into a wall.

Exactly, since 135 = 90 + 45. Mathematically the gradient is related by m = tan(θ), so you can remember that for next time. Often the greatest difficulty comes in trying to visualise the problem. Well done on getting it!

 

[johnstobbart]

Thanks a lot, JesseC, dydxforsn and Ray Vickson. I really appreciate the time and effort you took to help me solve this problem. These forums are a huge help. Thanks a lot!