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

This image contains two calculus problems.

Problem 4 asks to find the points on a parametric curve where the tangent line is horizontal. This involves finding where the derivative dy/dx is equal to zero.

Problem 5 asks to set up, but not evaluate, an integral to find the volume of a solid of revolution. The region is bounded by the curves y = x(x-3)^2 and y = 0, and it is rotated about the line x = 5. You'd need to use either the disk/washer method or the shell method to set up the integral.

Answer

4. $(6, -16)$ and $(2, 16)$ 5. $V = 2\pi \int_0^3 (5-x)(x(x-3)^2) \, dx$

Steps to Solve

Find $\frac{dy}{dt}$

Given $y = t^3 - 12t$, differentiate with respect to $t$: $$ \frac{dy}{dt} = 3t^2 - 12 $$

Find $\frac{dx}{dt}$

Given $x = t^2 + t$, differentiate with respect to $t$: $$ \frac{dx}{dt} = 2t + 1 $$

Find $\frac{dy}{dx}$

Using the chain rule, $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 12}{2t + 1} $$

Set $\frac{dy}{dx} = 0$ to find horizontal tangents

A horizontal tangent occurs when $\frac{dy}{dx} = 0$, which means the numerator must be zero while ensuring the denominator is not also zero (as that would give an indeterminate form).

$$ 3t^2 - 12 = 0 $$ $$ 3(t^2 - 4) = 0 $$ $$ t^2 = 4 $$ $$ t = \pm 2 $$

Check that denominator is not equal to zero

Check if $2t + 1 = 0$ for $t = \pm 2$. 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 $(x, y)$

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 the volume of the solid of revolution

The region is bounded by $y = x(x-3)^2$ and $y=0$, rotated about $x = 5$.

Notice that we are rotating around the line $x=5$ which is a vertical line, so we should integrate with respect to $x$. Also, we need to use the shell method since we are rotating around a vertical line and integrating with respect to $x$.

The shell method formula is $V = 2\pi \int_a^b r(x)h(x) , dx$, where

$r(x)$ is the radius of the shell which is $5-x$
$h(x)$ is the height of the shell, which is $x(x-3)^2$
$a$ and $b$ are the $x$-intercepts where $y=x(x-3)^2$ intersects $y=0$, which occur at $x = 0$ and $x = 3$.

The integral for the volume is: $$ V = 2\pi \int_0^3 (5-x)(x(x-3)^2) , dx $$

$(6, -16)$ and $(2, 16)$
$V = 2\pi \int_0^3 (5-x)(x(x-3)^2) , dx$

More Information

The points where the tangent line is horizontal are $(6, -16)$ and $(2, 16)$. The volume of the solid of revolution is given by the integral $2\pi \int_0^3 (5-x)(x(x-3)^2) , dx$. We were only asked to set this up, not evaluate it.

Tips

Forgetting to check if the denominator $\frac{dx}{dt}$ is zero when $\frac{dy}{dx} = 0$.
Using the disk/washer method instead of the shell method when rotating about a vertical line and integrating with respect to $x$.
Incorrectly determining the limits of integration.
Incorrectly setting up the radius or height in the volume integral.

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