1. a.) Find an equation for the line perpendicular to the tangent curve y=x^3 - 9x + 5 at the point (3,5)
[* for a. the answer that I obtained was
y-5 = -1/18 (x-3) ]
b.) What is the smallest slope on the curve? At what point on the curve does the curve have this slope?
c.) Find equations for the tangents to the curve at the points where the slope of the curve is 18.
a) correct
b) smallest slope ---> minimum slope
so set derivative of slope = 0
that is, the 2nd derivative = 0 and solve
c) set 3x^2 - 9 = 18 and solve
You will get two differenent x's, find the matching y's and do the usual thing to find the equation.
a.) Oh, nice! It looks like you've already found the equation for the line perpendicular to the tangent curve at the point (3,5). Great job!
b.) Ah, the smallest slope on the curve. Well, imagine a snail going uphill. It's probably going pretty slow, right? Similarly, the smallest slope on the curve would correspond to the flattest part, where the snail is moving at its snail pace. To find this point, we need to find where the derivative of the curve equals zero.
c.) Alright, let's find the equations of the tangents to the curve where the slope is 18. You could call these tangents "slope mates" since they have something in common with each other. To find these points, we'll need to find where the derivative of the curve equals 18. Hang in there, we're getting closer!
a) To find the equation of a line perpendicular to the tangent curve at the point (3,5), we need to determine the slope of the tangent curve at that point, and then find the negative reciprocal of that slope.
To find the slope of the tangent curve at (3,5), we can take the derivative of the curve y = x^3 - 9x + 5 and evaluate it at x = 3.
dy/dx = 3x^2 - 9
Evaluate at x = 3:
dy/dx = 3(3)^2 - 9 = 3(9) - 9 = 18 - 9 = 9
So, the slope of the tangent curve at (3,5) is 9.
The negative reciprocal of 9 is -1/9. Therefore, the slope of the line perpendicular to the tangent curve is -1/9.
Now, we can use the point-slope form of the equation to find the equation of the line:
y - y1 = m(x - x1)
Using point (3,5) and slope -1/9, the equation becomes:
y - 5 = (-1/9)(x - 3)
Simplifying,
y - 5 = (-1/9)x + 1/3
Rearranging,
y = (-1/9)x + 1/3 + 5
y = (-1/9)x + 1/3 + 15/3
y = (-1/9)x + 16/3
Therefore, the equation of the line perpendicular to the tangent curve y = x^3 - 9x + 5 at the point (3,5) is y = (-1/9)x + 16/3.
b) To find the smallest slope on the curve, we need to find the minimum value of the derivative dy/dx.
Take the derivative of the curve y = x^3 - 9x + 5:
dy/dx = 3x^2 - 9
The smallest slope occurs at the minimum or maximum point of the curve, which can be found by setting dy/dx equal to zero:
3x^2 - 9 = 0
3x^2 = 9
x^2 = 9/3
x^2 = 3
Taking the square root of both sides:
x = ±√3
Since we are looking for the smallest slope, we consider the negative square root:
x = -√3
To find the smallest slope, we substitute this value of x back into the derivative:
dy/dx = 3(-√3)^2 - 9
= 3(3) - 9
= 9 - 9
= 0
Therefore, the smallest slope on the curve is 0. The point on the curve where the slope is 0 is when x = -√3.
c) To find equations for the tangents to the curve at the points where the slope of the curve is 18, we need to find the x-values where dy/dx = 18. Then, for each x-value, we can find the corresponding y-value to determine the point of tangency.
Setting dy/dx = 18:
3x^2 - 9 = 18
3x^2 = 27
x^2 = 9
x = ±3
For x = 3:
Substituting x = 3 into the equation y = x^3 - 9x + 5:
y = (3)^3 - 9(3) + 5
y = 27 - 27 + 5
y = 5
So, the first point of tangency is (3,5).
For x = -3:
Substituting x = -3 into the equation y = x^3 - 9x + 5:
y = (-3)^3 - 9(-3) + 5
y = -27 + 27 + 5
y = 5
So, the second point of tangency is (-3,5).
The equations of the tangents to the curve at these points can be found using the point-slope form:
For (3,5):
Using the slope 18, the equation becomes:
y - 5 = 18(x - 3)
Simplifying,
y - 5 = 18x - 54
y = 18x - 49
For (-3,5):
Using the slope 18, the equation becomes:
y - 5 = 18(x + 3)
Simplifying,
y - 5 = 18x + 54
y = 18x + 59
Therefore, the equations of the tangents to the curve at the points where the slope is 18 are y = 18x - 49 and y = 18x + 59.
a.) To find the equation for the line perpendicular to the tangent curve at the point (3,5), we need to find the slope of the tangent curve at that point, and then find the negative reciprocal of this slope, as perpendicular lines have slopes that are negative reciprocals of each other.
1. First, find the derivative of the original curve y = x^3 - 9x + 5 with respect to x to get the slope function: dy/dx = 3x^2 - 9.
2. Evaluate the derivative at x = 3 to find the slope of the tangent line at that point: dy/dx = 3*(3^2) - 9 = 18 - 9 = 9.
3. The negative reciprocal of 9 is -1/9, so the slope (m) of the line perpendicular to the tangent curve is -1/9.
4. Now we can use the point-slope form of a line to find the equation of the line: y - y1 = m(x - x1), where (x1, y1) is the given point (3,5).
Plugging in the values, we get: y - 5 = (-1/9)(x - 3)
b.) To find the smallest slope on the curve, we need to find the minimum value of the derivative function. The minimum value occurs at the x-coordinate of the vertex of the original curve.
1. Find the derivative of the original curve y = x^3 - 9x + 5: dy/dx = 3x^2 - 9.
2. Set the derivative equal to zero and solve for x to find the x-coordinate of the vertex: 3x^2 - 9 = 0. Simplifying, we get: x^2 - 3 = 0. Factoring, we have: (x - √3)(x + √3) = 0. Solving for x, we get: x = √3 or x = -√3.
3. Since the curve is symmetric, the vertex is located at the x-coordinate where the derivative is zero, so we choose x = √3 as the x-coordinate of the vertex.
4. Evaluate the original curve at x = √3 to find the y-coordinate of the vertex: y = (√3)^3 - 9(√3) + 5.
c.) To find the equation of the tangent lines at the points where the slope of the curve is 18, we need to find the x-values where the derivative is equal to 18 and then find the corresponding y-values.
1. Set the derivative function equal to 18 and solve for x: 3x^2 - 9 = 18.
2. Rearrange the equation and divide by 3: x^2 - 9 = 6. Simplifying, we get: x^2 = 15. Taking the square root, we have: x = ±√15.
3. Evaluate the original curve at x = √15 and x = -√15 to find the corresponding y-values for the points where the slope is 18: y = (√15)^3 - 9(√15) + 5 and y = (-√15)^3 - 9(-√15) + 5.
Therefore, the equations for the tangents to the curve at the points where the slope is 18 are:
1. y = (√15)^3 - 9(√15) + 5
2. y = (-√15)^3 - 9(-√15) + 5