1. Sketch the curves y=e^x and y=e^-2x, using the same axes. The line y=4 intersects the first curve at A and the second curve at B. Calculate the length AB to two decimal places.
2. Find the coordinates of the turning point on the curve y=2e^3x+8e^-3x and determine the nature of this turning point.
For 1, I have sketched the curves, but i'm not sure how to find the length. I thought I could do y=y to find the x co-ordinates (and use those as my limits) for each but I don't know how to find them out with e^x.
For 2, I differentiated it to get y=6e^3x-24e^-3x and don't know what to do after that.
Please help. Thanks in advance for any help!
2. find each x by
4=e^x
ln 4= x so the point is ln4,4
and the other point is
ln2=-2x
x=-ln4/2 and the boint B is -1/2 ln4,4
Use the distance formula to find the length.
1. Since the y coordinates of A and B are the same, the distance between the points will be the difference in the X values of those two points.
2. The derivative is
y' = 6e^3x -24e^-3x
When that equals zero,
e^3x = 4e^-3x
e^6x = 4
6x = ln 4 = 1.3863
x = 0.23105 is an extreme point of the function. The value of y there is
2 e^0.693 + 8 e^-0.693 = 2*2 + 8/(1/2) = 20
Take the second derivative of y(x) to see if (0.23105,20) is a minumum or maximum. It is going to be positive. You also see that, since y becomes infinite as x gets much larger, it must be a minimum.
y''(x) =
For question 1, to find the length AB of the curve y = e^x and y = e^(-2x) between the intersection points A and B, you can use the arc length formula:
L = ∫ sqrt(1 + (dy/dx)^2) dx
Step 1: Find the intersection points A and B by setting y = 4 for each curve and solving for x.
For the first curve, y = e^x, set e^x = 4 and solve for x. Taking the natural logarithm of both sides, you get x = ln(4).
For the second curve, y = e^(-2x), set e^(-2x) = 4 and solve for x. Again, taking the natural logarithm, you get -2x = ln(4), and dividing by -2, x = -0.5 ln(4).
So the x-coordinate of point A is ln(4), and the x-coordinate of point B is -0.5 ln(4).
Step 2: Calculate the length using the arc length formula.
L = ∫[x=a to b] sqrt(1 + (dy/dx)^2) dx
For the first curve, y = e^x, the derivative is dy/dx = e^x.
For the second curve, y = e^(-2x), the derivative is dy/dx = -2e^(-2x).
Now substitute these derivatives into the arc length formula:
L = ∫[x=a to b] sqrt(1 + (dy/dx)^2) dx
L = ∫[x=ln(4) to -0.5ln(4)] sqrt(1 + (e^x)^2) dx + ∫[x=ln(4) to -0.5ln(4)] sqrt(1 + (-2e^(-2x))^2) dx
Numerically evaluating these integrals will give you the length of AB.
For question 2:
To find the turning point on the curve y = 2e^(3x) + 8e^(-3x), you need to find the derivative of y and set it equal to 0.
Step 1: Take the derivative of y with respect to x.
y' = 6e^(3x) - 24e^(-3x)
Step 2: Set y' equal to 0 and solve for x.
6e^(3x) - 24e^(-3x) = 0
Divide both sides by 6 to simplify the equation:
e^(3x) - 4e^(-3x) = 0
Step 3: Let u = e^(3x), then the equation becomes u - 4/u =0.
Multiply both sides by u to eliminate the denominator:
u^2 - 4 = 0
Factor the equation:
(u - 2)(u + 2) = 0
Step 4: Solve for u.
u - 2 = 0 or u + 2 = 0
u = 2 or u = -2
Step 5: Substitute back u = e^(3x) into the equation to find x.
For u = 2: e^(3x) = 2
Take the natural logarithm of both sides: ln(e^(3x)) = ln(2)
3x = ln(2)
x = ln(2) / 3
For u = -2: e^(3x) = -2 (This doesn't have a real solution because e^(3x) is always positive)
So the turning point on the curve y = 2e^(3x) + 8e^(-3x) is at (ln(2) / 3, y(ln(2) / 3)), where y(ln(2) / 3) can be found by substituting x = ln(2) / 3 into the original equation.
To determine the nature of the turning point, you can examine the second derivative of y at that point.