1. Find the points on the parametric curve x = t^2 + t, y = t^3 - 12t, where the tangent is horizontal. 2. Set up, but DO NOT evaluate, an integral for the volume of the solid obta... 1. Find the points on the parametric curve x = t^2 + t, y = t^3 - 12t, where the tangent is horizontal. 2. Set up, but DO NOT evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the curves y = x(x - 3)^2 and y = 0 about the line x = 5.
Understand the Problem
The image contains two separate math questions. The first question asks to find the points on a parametric curve where the tangent is horizontal, which requires finding where the derivative dy/dx equals zero. The second part asks to set up, but not evaluate, an integral for the volume of a solid obtained by rotating a region bounded by two curves around a vertical line, which can be solved using the disk or shell method.
Answer
4. $(6, -16)$, $(2, 16)$ 5. $V = \int_{0}^{3} 2\pi (5-x) x(x-3)^2 dx$
Answer for screen readers
- The points where the tangent is horizontal are $(6, -16)$ and $(2, 16)$.
- The integral for the volume is $V = \int_{0}^{3} 2\pi (5-x) x(x-3)^2 dx$.
Steps to Solve
- Find $\frac{dy}{dt}$
Given $y = t^3 - 12t$, differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = 3t^2 - 12$
- Find $\frac{dx}{dt}$
Given $x = t^2 + t$, differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 2t + 1$
- Find $\frac{dy}{dx}$
Using the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Thus,
$\frac{dy}{dx} = \frac{3t^2 - 12}{2t + 1}$
- Find where $\frac{dy}{dx} = 0$
A horizontal tangent occurs where $\frac{dy}{dx} = 0$. This happens when the numerator is zero and the denominator is not zero. So, we need to solve $3t^2 - 12 = 0$ for $t$.
$3t^2 - 12 = 0$
$3t^2 = 12$
$t^2 = 4$
$t = \pm 2$
Check that the denominator is not zero at these values of $t$:
For $t = 2$, $2t + 1 = 2(2) + 1 = 5 \neq 0$. For $t = -2$, $2t + 1 = 2(-2) + 1 = -3 \neq 0$. So, both $t = 2$ and $t = -2$ are valid.
- Find the points $(x, y)$
Substitute $t = 2$ and $t = -2$ into the equations for $x$ and $y$.
For $t = 2$: $x = (2)^2 + 2 = 4 + 2 = 6$ $y = (2)^3 - 12(2) = 8 - 24 = -16$ So, the point is $(6, -16)$.
For $t = -2$: $x = (-2)^2 + (-2) = 4 - 2 = 2$ $y = (-2)^3 - 12(-2) = -8 + 24 = 16$ So, the point is $(2, 16)$.
- Set up the integral for volume
We are rotating the region bounded by $y = x(x-3)^2$ and $y = 0$ about the line $x = 5$. The curve $y = x(x-3)^2 = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x$. The curve intersects $y=0$ when $x(x-3)^2 = 0$, which means $x = 0$ or $x = 3$. The region is bounded between $x = 0$ and $x = 3$.
Since the axis of rotation is vertical, we can use the method of cylindrical shells. The radius of a shell is the distance from the axis of rotation to $x$, which is $5 - x$. The height of the shell is the value of $y = x(x-3)^2$. The thickness of the shell is $dx$. So, the volume of a shell is $2\pi(5-x)(x(x-3)^2)dx$.
Therefore, the integral for the volume is: $V = \int_{0}^{3} 2\pi (5-x) x(x-3)^2 dx$
- The points where the tangent is horizontal are $(6, -16)$ and $(2, 16)$.
- The integral for the volume is $V = \int_{0}^{3} 2\pi (5-x) x(x-3)^2 dx$.
More Information
The cylindrical shell method is particularly useful when rotating about a vertical line when the function is defined in terms of x versus y. The formula to find the volume using the shell method when rotating around the y-axis is expressed as: $V = 2\pi \int_{a}^{b} x f(x) dx$
Tips
- Forgetting to check that $\frac{dx}{dt} \neq 0$ when $\frac{dy}{dt} = 0$.
- Setting up the volume integral incorrectly by using the wrong radius or limits of integration. A common mistake is to use $x$ as the radius instead of the distance from the axis of rotation, which in this case is $5 - x$.
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