Rate of Change in X-intercept

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turophile

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The problem:

A point Q starts at Q[sub:i773cbkc]0[/sub:i773cbkc] = (1, 0) and moves along the upper semicircle x[sup:i773cbkc]2[/sup:i773cbkc] + y[sup:i773cbkc]2[/sup:i773cbkc] = 1 with dx/dt = – 2. When x = 1/sqrt(2), at what rate is the x-intercept of the tangent line to the curve at Q increasing?

My work so far:

The slope of the tangent line is dy/dx where y = (1 - x[sup:i773cbkc]2[/sup:i773cbkc])[sup:i773cbkc]1/2[/sup:i773cbkc].

dy/dx = d[(1 – x[sup:i773cbkc]2[/sup:i773cbkc])[sup:i773cbkc]1/2[/sup:i773cbkc]]/dx
= 1/2 · (1 – x[sup:i773cbkc]2[/sup:i773cbkc])[sup:i773cbkc]- 1/2[/sup:i773cbkc] · d(1 – x[sup:i773cbkc]2[/sup:i773cbkc])/dx
= 1/2 · (1 – x[sup:i773cbkc]2[/sup:i773cbkc])[sup:i773cbkc]- 1/2[/sup:i773cbkc] · (– 2x)
= – x · (1 – x[sup:i773cbkc]2[/sup:i773cbkc])[sup:i773cbkc]- 1/2[/sup:i773cbkc]

When x = 1/sqrt(2), y = 1/sqrt(2), and the slope of the tangent line is - 1.

The equation of the tangent line is y - 1/sqrt(2) = (- 1)(x - 1/sqrt(2). The x-intercept is the solution to this equation when y = 0.

I'm not sure where to go next with this one.
 
turophile said:
The problem:

A point Q starts at Q[sub:3qyxw5q1]0[/sub:3qyxw5q1] = (1, 0) and moves along the upper semicircle x[sup:3qyxw5q1]2[/sup:3qyxw5q1] + y[sup:3qyxw5q1]2[/sup:3qyxw5q1] = 1 with dx/dt = – 2. When x = 1/sqrt(2), at what rate is the x-intercept of the tangent line to the curve at Q increasing?

My work so far:

The slope of the tangent line is dy/dx where y = (1 - x[sup:3qyxw5q1]2[/sup:3qyxw5q1])[sup:3qyxw5q1]1/2[/sup:3qyxw5q1].

dy/dx = d[(1 – x[sup:3qyxw5q1]2[/sup:3qyxw5q1])[sup:3qyxw5q1]1/2[/sup:3qyxw5q1]]/dx
= 1/2 · (1 – x[sup:3qyxw5q1]2[/sup:3qyxw5q1])[sup:3qyxw5q1]- 1/2[/sup:3qyxw5q1] · d(1 – x[sup:3qyxw5q1]2[/sup:3qyxw5q1])/dx
= 1/2 · (1 – x[sup:3qyxw5q1]2[/sup:3qyxw5q1])[sup:3qyxw5q1]- 1/2[/sup:3qyxw5q1] · (– 2x)
= – x · (1 – x[sup:3qyxw5q1]2[/sup:3qyxw5q1])[sup:3qyxw5q1]- 1/2[/sup:3qyxw5q1]

When x = 1/sqrt(2), y = 1/sqrt(2), and the slope of the tangent line is - 1.

The equation of the tangent line is y - 1/sqrt(2) = (- 1)(x - 1/sqrt(2). The x-intercept is the solution to this equation when y = 0.

I'm not sure where to go next with this one.
Click to expand...

So how are the dx/dt and dy/dt are related?

Try to visualize the problem - as the point is moving the equation of the tangent line is changing - now what happens?
 
y - 1/sqrt(2) = (- 1)(x - 1/sqrt(2) ? dy/dt = - dx/dt = - (- 2) = 2

Since dx/dt is - 2 and dy/dt = 2, the x-intercept of the tangent line is increasing. I don't see how to calculate its rate of change, though.
 
Try thinking Pythagoras. The tangent line forms a triangle from where it is tangent to where it crosses the x-axis.

One way to think about it.
 
If I am following correctly, the triangle I visualize is the right triangle ABC with vertices A = (1/sqrt[2], 1/sqrt[2]), B = (1/sqrt[2], 0), and C = (q, 0), where q is the x-intercept. If we let a = AC, b = AB, and c = AC, then a[sup:9xfv0u52]2[/sup:9xfv0u52] + b[sup:9xfv0u52]2[/sup:9xfv0u52] = c[sup:9xfv0u52]2[/sup:9xfv0u52]. We want to find dq/dt. Is that good so far?
 
Take a general point (x,y)on the semi-circle and let the intercept of the tangent = b (in this case 'b' is not constant - it is a function of x & y)

then using Pythagorus

[(b-x)[sup:movk80tw]2[/sup:movk80tw] + y[sup:movk80tw]2[/sup:movk80tw]] + [1[sup:movk80tw]2[/sup:movk80tw]] = b[sup:movk80tw]2[/sup:movk80tw]

now continue...
 
I'm not sure I understand how the Pythagorean Theorem is being applied here. I can see that (b-x)[sup:32h2qz4h]2[/sup:32h2qz4h] + y[sup:32h2qz4h]2[/sup:32h2qz4h] is equal to the square of the distance from (x, y) to (b, 0). How do I get from there to [(b - x)[sup:32h2qz4h]2[/sup:32h2qz4h] + y[sup:32h2qz4h]2[/sup:32h2qz4h]] + [1[sup:32h2qz4h]2[/sup:32h2qz4h]] = b[sup:32h2qz4h]2[/sup:32h2qz4h]?
 
\(\displaystyle f(x)=\sqrt{1-x^{2}}\)

\(\displaystyle f'(x)=\frac{-x}{\sqrt{1-x^{2}}}\)

Let the point on the x-axis the line passes through be (b,0).

The equation of the line is

\(\displaystyle y=f'(x)(x-b)+f(x)\)

\(\displaystyle 0=\frac{-x}{\sqrt{1-x^{2}}}(x-b)+\sqrt{1-x^{2}}\)

\(\displaystyle b=\frac{2x^{2}-1}{x}\)

Now, can you finish up?.
 
db/dt = d/dt [(2x[sup:2w51r2ko]2[/sup:2w51r2ko] - 1)/x]

= d/dt (2x - 1/x)

= 2 * dx/dt + 1/x[sup:2w51r2ko]2[/sup:2w51r2ko] * dx/dt

= (2 + 1/x[sup:2w51r2ko]2[/sup:2w51r2ko]) * dx/dt

= (2 + 1/[1/sqrt(2)][sup:2w51r2ko]2[/sup:2w51r2ko]) * dx/dt

= (2 + 2) * (- 2)

= 4 * (- 2)

= - 8

That's counterintuitive since the rate of change in the x-intercept should be positive. Also, my textbook gives the answer as 4. I'm still stuck, apparently.
 
That's OK. You do not need to attach the dx/dt=-2. You get 4. This gives the change in b when x=1/sqrt(2). Which is db/dx.

But, we can get it SK's way as well.

\(\displaystyle (b-x)^{2}+y^{2}+1=b^{2}\)

\(\displaystyle b=\frac{1}{2}x+\frac{y^{2}}{2x}+\frac{1}{2x}\)

Now, the implicit differentiation is rather icky, but not too bad.

\(\displaystyle \frac{db}{dt}=\frac{1}{2}\cdot \frac{dx}{dt}+\frac{\left(2x(2y\cdot \frac{dy}{dt})-y^{2}\cdot 2\cdot \frac{dx}{dt}\right)}{4x^{2}}+\frac{\left(2x(0)-1(2\cdot \frac{dx}{dt})\right)}{4x^{2}}\)

Now, letting \(\displaystyle dx/dt=-2, \;\ dy/dt=2, \;\ x=\frac{1}{\sqrt{2}}, \;\ y=\frac{1}{\sqrt{2}}\) we get ...............drum roll................\(\displaystyle \boxed{\frac{db}{dt}=4}\)

I thought we may as well finish this thing up. It has languished long enough. :D
 
Galactus missed a simplification that would make the differentiation very easy..

\(\displaystyle (b-x)^2 \ \ + \ \ y^2 + 1^2 \ \ = \ \ b^2\)

\(\displaystyle b^2 \ \ - \ \ 2bx + \ \ x^2 \ \ + \ \ y^2 + 1 \ \ = \ \ b^2\)

we know x[sup:29eafuq7]2[/sup:29eafuq7] + y[sup:29eafuq7]2[/sup:29eafuq7] = 1 .... then...

\(\displaystyle - \ \ 2bx + \ \ 1 \ \ + \ \ 1 \ \ = \ \ b^2 \ \ - \ \ b^2\)

\(\displaystyle 2bx \ \ = \ \ 2\)

\(\displaystyle b \ \ = \ \ \frac{1}{x}\)

\(\displaystyle \frac{db}{dt} \ \ = \ \ - \ \ \frac{1}{x^2} \cdot \frac{dx}{dt}\)

and we are there....
 
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