At what point of the curve y= x^4 is the normal line parallel to 2x + y = 3? Find the equation of the normal line.
the normal line is perpendicular to the tangent
the slope of the tangent is the 1st derivative
normal slope ... -2
tangent slope ... 1/2
1/2 = 4 x^3
... 1/8 = x^3
... 1/2 = x
point ... (1/2 , 1/16)
equation ... y - 1/16 = -2 (x - 1/2)
Oh, normal lines and parallelism, we're getting fancy here! Hang on tight, because I'm about to clown around with some math.
To find the point on the curve y = x^4 where the normal line is parallel to 2x + y = 3, we need to find the slope of the normal line and equate it with the slope of the given line.
First, let's find the slope of the given line. It's in the form of y = mx + b, where m is the slope. We can rewrite it as y = -2x + 3. So the slope is -2.
Now, the slope of the normal line is the negative reciprocal of the slope of the tangent line to the curve at the point we're looking for.
The derivative of y = x^4 is dy/dx = 4x^3. To find the slope of the tangent line at any given point (x, y), we substitute x into the derivative.
Let's call the x-coordinate of the point we're looking for as "a". So, dy/dx at x = a is 4a^3.
To find the slope of the normal line, we take the negative reciprocal of 4a^3. That is, -1/(4a^3).
Now, we equate this slope (-1/(4a^3)) with the slope of the given line (-2). Time for some equation-solving circus act!
-1/(4a^3) = -2
Multiplying both sides by (4a^3), we get:
-1 = -8a^3
Dividing by -8, we have:
1/8 = a^3
Taking the cube root of both sides, we find:
a = (∛(1/8))
Simplifying this, we get:
a = 1/2
So, the point on the curve y = x^4 where the normal line is parallel to 2x + y = 3 is (1/2, 1/16).
Finally, let's find the equation of the normal line. The point-slope form of a line is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, and m is the slope of the line.
Using (1/2, 1/16) as the point and the slope we found earlier (-1/(4a^3) = -1/(4(1/2)^3) = -1/8), we can write the equation of the normal line as:
y - 1/16 = (-1/8)(x - 1/2)
Now, let me put on my silly clown hat and simplify this:
16y - 1 = -2x + 1
Rewriting it:
2x + 16y = 2
There you have it! The equation of the normal line is 2x + 16y = 2.
To find the point on the curve y = x^4 where the normal line is parallel to 2x + y = 3, we first need to find the slope of the normal line.
The equation 2x + y = 3 can be rearranged to the slope-intercept form, y = -2x + 3. Comparing this with the general equation of a line, y = mx + c, we can see that the slope of the line is -2.
For a curve y = x^4, we can find the derivative to find the slope of the tangent line at any point on the curve. The derivative of y = x^4 is dy/dx = 4x^3.
The slope of the normal line will be the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line at any given point on the curve y = x^4 is -1/(4x^3).
Since we want the normal line to be parallel to the line 2x + y = 3, we set the slopes equal to each other:
-1/(4x^3) = -2
Now, solve for x:
-1 = -8x^3
Divide both sides by -8:
1/8 = x^3
Take the cube root of both sides:
x = 0.5
Now we have the x-coordinate of the point on the curve. To find the y-coordinate, substitute x into the equation y = x^4:
y = (0.5)^4
y = 0.0625
Therefore, the point on the curve y = x^4 where the normal line is parallel to 2x + y = 3 is (0.5, 0.0625).
To find the equation of the normal line, we can use the point-slope form of a line. The equation of a line with slope m passing through a point (x1, y1) is given by y - y1 = m(x - x1).
Using the point (0.5, 0.0625) and the slope of the normal line -1/(4x^3) = -1/(4 * (0.5)^3) = -1/ (4 * 0.125) = -1/0.5 = -2, we can write the equation of the normal line as:
y - 0.0625 = -2(x - 0.5)
Simplifying:
y - 0.0625 = -2x + 1
Rearranging to the slope-intercept form, we get:
y = -2x + 1.0625
Therefore, the equation of the normal line is y = -2x + 1.0625.
To find the point on the curve where the normal line is parallel to the given line, we need to determine the slope of the normal line and then find the corresponding x-coordinate on the curve.
Step 1: Find the slope of the given line.
The given line is 2x + y = 3. To find its slope, we need to rearrange the equation into slope-intercept form (y = mx + b), where m is the slope.
Rewriting the equation in slope-intercept form:
y = -2x + 3
Comparing this equation with y = mx + b, we can see that the slope (m) of the given line is -2.
Step 2: Find the slope of the normal line.
The slope of the normal line would be the negative reciprocal of the slope of the given line. So, the slope of the normal line would be 1/2.
Step 3: Find the x-coordinate on the curve.
Now that we have the slope of the normal line, we need to find the point on the curve y = x^4 that corresponds to this slope.
Let's differentiate the equation y = x^4 with respect to x:
dy/dx = 4x^3
This gives us the equation of the derivative, which represents the slope of the curve at any given point.
Since the slope of the normal line is the negative reciprocal of the slope of the curve, we can set the two slopes equal to each other:
1/2 = 4x^3
Step 4: Solve for x.
Solving the equation 1/2 = 4x^3 for x will give us the x-coordinate(s) at which the normal line is parallel to 2x + y = 3.
1/2 = 4x^3
Divide both sides by 4:
1/8 = x^3
Taking the cube root of both sides:
x = (1/8)^(1/3)
Step 5: Find the corresponding y-coordinate.
To find the corresponding y-coordinate, we substitute the x-value obtained from Step 4 into the equation y = x^4:
y = ((1/8)^(1/3))^4
Step 6: Simplify the y-coordinate.
Calculating ((1/8)^(1/3))^4 will give us the y-coordinate on the curve.
The equation of the normal line through this point can be determined using the point-slope formula or by finding the equation of the line that passes through this point with a slope of 1/2.