how do i solve this problem?
find an equation for a line that is tangent to the graph of y=e^x and goes through the origin.
let the point of contact be (a,e^a)
so the lope of the tangent is (e^a - 0)/(a-0) = e^a/a
also y' = e^x
at (a,e^a) , y' = e^a
so e^a/a = e^a
then a = 1
and the slope of the tangent is e^1 = e
and the eqution of the tangent, using y = mx+b, is
y = ex + b
but the y=intercept is zero, it goes through the origin
our equation of the tangent is y = ex
Well, solving this problem is no exponential task! To find an equation for a line that is tangent to the graph of y=e^x and goes through the origin, we need to determine the slope of that tangent line.
The derivative of y=e^x is simply e^x. So, at any given point (x, y) on the graph, the slope of the tangent line will be e^x. Since we want the line to go through the origin, which is (0, 0), we have our y-intercept.
So, we have the equation of our line: y = e^x * x.
But hey, don't worry! The calculation is not as complicated as it seems. Just remember that humor and math always add up to a great solution!
To find the equation of a line that is tangent to the graph of y=e^x and passes through the origin, you can follow these steps:
Step 1: Find the slope of the tangent line at the point of tangency.
- The derivative of y=e^x is dy/dx = e^x.
- Since the tangent line needs to pass through the same point on the graph, we evaluate dy/dx at that point.
- At x=0, dy/dx = e^0 = 1. So, the slope of the tangent line is 1.
Step 2: Determine the y-intercept of the tangent line.
- Since the line passes through the origin (0,0), the y-intercept is 0.
Step 3: Write the equation of the tangent line using the slope-intercept form (y = mx + b).
- Substitute the slope and y-intercept into the equation obtained in step 2.
- The equation is y = 1x + 0.
- Simplify it to y = x.
Therefore, the equation of the line that is tangent to the graph of y=e^x and passes through the origin is y = x.
To solve this problem, we need to find the equation of the tangent line to the graph of y = e^x that passes through the origin (0, 0).
Step 1: Find the derivative of the function y = e^x.
The derivative of y = e^x is dy/dx = e^x.
Step 2: Find the slope of the tangent line.
Since the derivative gives the slope of the tangent line at any point on the graph, we can substitute the x-coordinate of the point of tangency into the derivative. Since the point of tangency is not given, we can denote it as (a, e^a) since it lies on the graph of y = e^x.
Substituting a into the derivative dy/dx = e^x, we get the slope of the tangent line at (a, e^a) as m = e^a.
Step 3: Use the point-slope form to find the equation of the tangent line.
Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
Since the point (a, e^a) lies on the line, we can substitute a for x1 and e^a for y1. The slope, e^a, can be substituted for m. Thus, the equation becomes:
y - e^a = e^a(x - a)
Step 4: Substitute the origin (0, 0) into the equation to find the value of a.
Since the equation of the tangent line passes through the origin, we can substitute x = 0 and y = 0 into the equation:
0 - e^a = e^a(0 - a)
- e^a = - a * e^a
Step 5: Solve for a.
We can divide both sides of the equation by -e^a, which gives:
1 = a
Therefore, a = 1.
Step 6: Substitute the value of a back into the equation to get the final equation of the tangent line.
Substituting a = 1 into the equation found in Step 3, we have:
y - e = e(x - 1)
Hence, the equation for a line that is tangent to the graph of y = e^x and passes through the origin is y - e = e(x - 1).