Implicit Differentiation

[ohai]

If someone can help me out here, that would be fab..

1. Find the two points where the curve x^2 + xy + y^2 = 7 crosses the x axis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents?

2. The velocity of a falling body is v = 8 [(s-t)^(1/2)] +1 feet per second at the instant t (sec) the body has fallen s feet from its starting point. Show that the body's acceleration is 32 ft/(sec^2).

3. The line that is mornal to the curve x^2 + 2xy - 3(y)^2 = 0 at (1,1) intersects the curve at what other point?

Much thanks!

 

[Unco]

1. i) x-intecepts occur where y=0.
ii) Differentiate implicitly and rearrange to get dy/dx =... Plug the coordinate s of the x-intercepts into this to get the gradient of the tangents at these points.

3. i) Again, differentiate implicitly to get the gradient at of the tangent at (1,1).
ii) The normal at this point has a gradient equal to the negative reciprocal of the gradient of the tangent.
iii) Use the straight-line equation with m of the normal and the point (1,1) to find the line equation y = ...
iv) Plug this into the curve's equation to find the other point.

 

[soroban]

Hello, ohai!

I'll walk you through the first one . . .

1. Find the two points where the curve x² + xy + y² = 7 crosses the x axis,

and show that the tangents to the curve at these points are parallel.

What is the common slope of these tangents?

As Unco pointed out, x-intercepts occur when y = 0.

We have: .x² + x(0) + 0² .= .7
. . . . . . . . . . . . . . . _
. . Hence: .x .= .± √7
. . . . . . . . . . . . . . . . . . . _ . . . . . _
The x-intercepts are: .(√7,0), (-√7,0)


Differentiate implicitly: .2x + xy' + y + 2yy' .= .0

. . . . . . xy' + 2yy' .= .- 2x - y

. . . . . . (x + 2y)y' .= .-(2x + y)

. . . . . . . . . . . . . . . . . . 2x + y
. . . . . . . . . . . . y' .= .- ---------
. . . . . . . . . . . . . . . . . . x + 2y

. . . . . . . . . . . . . . . . . _
. . . . _ . . . . . . . . . 2(√7) + 0
At (√7,0): . y' .= .- ------------ . = . - 2
. . . . . . . . . . . . . . .√7 + 2(0)

. . . . . . . . . . . . . . . . . . . _
. . . . ._. . . . . . . . . . 2(-√7) + 0
At (-√7,0): , y' .= .- -------------- . = . - 2
. . . . . . . . . . . . . . . -√7 + 2(0)