In this problem we consider drawing some straight lines which form a nice pattern. Consider joining the point (0.1,0) to the point (0,0.9) by a line segment; then
joining (0.2,0) to (0,0.8) by a line segment; and so on. In general, consider joining the points (a,0) and (0,b) where a + b = 1 and 0 < a, b < 1.
We will show in this problem that all these straight lines are tangent to the curve with equation √x + √y = 1
(with 0 < x, y < 1).
Now i've done A and B but they may be useful as part of the others. I want to find out C and D but am having trouble can someone please help.
(a) Show that the line joining (a, 0) and (0, b), where a + b = 1 and 0 < a, b < 1, has equation
bx + ay = ab
(b) Show that if x and y are related by the equation
√x + √y = 1 then
dy
dx= -√y/x
(c) Let (p,q) be a point on the curve √x + √y = 1. Show that the tangent line to this curve at (p,q) has equation x*√q + y√p = √p*q
(d) Explain why the lines considered in parts (a) and (c) are the same, i.e. the lines we originally drew in part (a) are precisely the tangent lines to the curve
√x + √y=1
Thanks.
Hi again. Just realised that i got B wrong. So it means i need that too. Just to clarify (b) Show that if x and y are related by the equation
√x + √y = 1 then
dy _____
dx= -√(y/x)
_ _
I got -√y/-√x.
To solve part (c), you need to find the equation of the tangent line to the curve √x + √y = 1 at the point (p, q).
To find the equation of a tangent line, you need two pieces of information: the slope of the tangent line and a point it passes through.
First, let's find the slope of the tangent line. We can do this by taking the derivative of the equation √x + √y = 1 with respect to x.
Differentiating both sides of the equation with respect to x gives us:
(1/2)*1/√x + (1/2)*1/√y * (dy/dx) = 0
Now, let's solve for dy/dx, which represents the slope of the tangent line:
(1/2)*1/√y * (dy/dx) = - (1/2)*1/√x
(dy/dx) = -(√x/√y)
Next, we need a point on the tangent line. Since the tangent line passes through (p, q), we can substitute these values into the equation of the tangent line.
Using the point-slope form:
(y - q) = (dy/dx)*(x - p)
Substituting the values of dy/dx and the point (p, q) gives us:
(y - q) = - (√p/√q)*(x - p)
y - q = - (√p/√q)*x + (√p)*p
Now, rearranging the equation to the standard form:
x*√q + y*√p = √p*q
So, the equation of the tangent line to √x + √y = 1 at the point (p, q) is x*√q + y*√p = √p*q.
Now, to solve part (d), we need to explain why the lines considered in parts (a) and (c) are the same, i.e., the lines we originally drew in part (a) are precisely the tangent lines to the curve √x + √y = 1.
In part (a), we derived the equation of the line joining the points (a, 0) and (0, b) as bx + ay = ab.
Comparing this equation with the equation of the tangent line from part (c) (x*√q + y*√p = √p*q), we can see that they are the same equation.
To see this, set bx + ay = ab to x*√q + y*√p = √p*q. Since a + b = 1 and p + q = 1, we can substitute b = 1 - a and q = 1 - p into the equation.
This gives us x*√(1 - p) + y*√p = √(p * (1 - p)), which simplifies to x*√q + y*√p = √p*q.
Hence, the lines considered in parts (a) and (c) are the same, meaning the lines we originally drew in part (a) are precisely the tangent lines to the curve √x + √y = 1.