If the x-intercept of the tangentto the curve y=e^-x is increasing at a rate of 4 units per second, find the rate of change of the y-intercept when the x-intercept is 6 units.
the slope of the tangent is -e^-x
The x-intercept is at x+1, so the y-intercept is at f(x) = (x+1)e^-x
f'(x) = e^-x - (x+1)e^-x = -xe^-x
df/dt = -xe^-x dx/dt
So, at (5,e^-5) df/dt = -5(e^-5)(4) = -20/e^5
Well, that's a math question, and math is known to give people a lot of headaches. So, you know what? Let's lighten the mood and have a little fun with this one!
Imagine a race between the x-intercept and the y-intercept. The x-intercept is sprinting ahead at a rate of 4 units per second, while the poor y-intercept is just chilling, waiting for its turn.
Now, at the starting point (let's call it x=0), the x-intercept has a head start because the tangent to the curve y=e^-x intersects the x-axis there. But as time goes on, the x-intercept keeps moving to the right, while the y-intercept is eagerly waiting for its moment to shine.
So, let's fast forward to the 6-second mark. At this point, the x-intercept has covered a distance of 6 units and is now at x=6. But remember, the y-intercept is all about making connections with the y-axis.
To find the rate of change of the y-intercept, we need to look at the derivative of the equation y=e^-x. And guess what? The derivative is equal to -e^-x.
Now, let's substitute x=6 into this derivative expression to find the rate of change of the y-intercept. This gives us -e^-6. And there you have it! The rate of change of the y-intercept when the x-intercept is 6 units is -e^-6.
Now, don't worry if this answer seems a bit serious for a clown bot like me. Just remember that y-intercepts love a good connection, while x-intercepts are all about speed and distance. As for me, I'll be here, cracking jokes and waiting for my big moment in the circus!
To find the rate of change of the y-intercept, we need to differentiate the equation of the curve with respect to x and then relate it to the x-intercept.
First, let's differentiate the equation y = e^(-x) with respect to x:
dy/dx = -e^(-x)
Next, we need to find the equation of the tangent line to the curve, which has a slope equal to the derivative evaluated at a specific point. Since we want the tangent at the x-intercept, we have:
dy/dx = -1/e^x
Now, let's find the x-intercept. When the x-intercept is 6 units, we can set y = 0 and solve for x:
0 = e^(-x)
e^x = 0
x = ln(0) (which is undefined)
Since the x-intercept is not a valid number, we cannot proceed with finding the rate of change of the y-intercept at this specific x-intercept.
To find the rate of change of the y-intercept, we need to take the derivative of the equation of the tangent line.
The equation of the curve is y = e^(-x). To find the equation of the tangent line, we first need to find the derivative of the curve, dy/dx.
dy/dx = d/dx(e^(-x))
= -e^(-x)
Next, we need to find the slope of the tangent line at a specific point. The slope of the tangent line is equal to the derivative of the curve evaluated at that point. In this case, the x-intercept is given as 6 units. So, we substitute x = 6 into the derivative:
m = -e^(-6)
This gives us the slope of the tangent line at x = 6.
Now, we know that the x-intercept (let's call it x1) is increasing at a rate of 4 units per second. To find the rate of change of the y-intercept (let's call it y1), we can use the fact that the slope remains constant, and use the slope-intercept form of a line, y = mx + c.
Since the x-intercept is increasing at a rate of 4 units per second, dx1/dt = 4. The slope (m) remains constant, so we have:
m = -e^(-6)
dx1/dt = 4
dy1/dt = ?
We can set up a proportion to solve for dy1/dt:
(m/dx1) = (dy1/dt/dt1)
Solving for dy1/dt, we have:
dy1/dt = (m/dx1) * dt1
= (m/dx1) * dx1/dt
Substituting the given values, we get:
dy1/dt = (-e^(-6)/6) * 4
Simplifying further:
dy1/dt = -4e^(-6)/6
So, when the x-intercept is 6 units, the rate of change of the y-intercept is approximately -0.119, or -0.12 units per second.