Question

week 6 group project

1. use the chain rule to find the derivative of each function.

a. $y=(x^{3}+1)^{5}$
b. $y = sqrt{3x^{3}-4}$
c. $y=sin x^{4}$
d. $y=(x^{2}+3)(2x^{5}+9)^{4}$
e. $y=sqrt3{-x^{2}+2}$
f. $y=sec^{2}5x$

1. find equations of all lines tangent to $x + y^{3}-y = 1$ at $x = 1$.
2. an observer is 20 m above the ground floor of a large hotel atrium looking at a glass enclosed elevator shaft that is 20 m horizontally from the observer. the angle of elevation of the elevator is the angle that the observers line of sight makes with the horizontal (it may be positive or negative). assuming that the elevator rises at a rate of 5 m/s, what is the rate of change of the angle of elevation when the elevator is 10 m above the ground? when the elevator is 40 m above the ground?

Explanation:

1a. For $y=(x^{3}+1)^{5}$

Let $u = x^{3}+1$, then $y = u^{5}$.
By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.
$\frac{dy}{du}=5u^{4}$ and $\frac{du}{dx}=3x^{2}$.
Substitute $u = x^{3}+1$ back in: $\frac{dy}{dx}=5(x^{3}+1)^{4}\cdot3x^{2}=15x^{2}(x^{3}+1)^{4}$.

1b. For $y=\sqrt{3x^{3}-4}=(3x^{3}-4)^{\frac{1}{2}}$

Let $u = 3x^{3}-4$, then $y = u^{\frac{1}{2}}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=9x^{2}$.
$\frac{dy}{dx}=\frac{1}{2}(3x^{3}-4)^{-\frac{1}{2}}\cdot9x^{2}=\frac{9x^{2}}{2\sqrt{3x^{3}-4}}$.

1c. For $y = \sin(x^{4})$

Let $u=x^{4}$, then $y=\sin(u)$.
$\frac{dy}{du}=\cos(u)$ and $\frac{du}{dx}=4x^{3}$.
$\frac{dy}{dx}=\cos(x^{4})\cdot4x^{3}=4x^{3}\cos(x^{4})$.

1d. For $y=(x^{2}+3)(2x^{5}+9)^{4}$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=x^{2}+3$, $u^\prime = 2x$ and $v=(2x^{5}+9)^{4}$.
For $v=(2x^{5}+9)^{4}$, let $t = 2x^{5}+9$, then $v=t^{4}$, $\frac{dv}{dt}=4t^{3}$ and $\frac{dt}{dx}=10x^{4}$, so $\frac{dv}{dx}=4(2x^{5}+9)^{3}\cdot10x^{4}=40x^{4}(2x^{5}+9)^{3}$.
$\frac{dy}{dx}=2x(2x^{5}+9)^{4}+(x^{2}+3)\cdot40x^{4}(2x^{5}+9)^{3}=2x(2x^{5}+9)^{3}(2x^{5}+9 + 20x^{5}+60x^{3})=2x(2x^{5}+9)^{3}(22x^{5}+60x^{3}+9)$.

1e. For $y=\sqrt[3]{-x^{2}+2}=(-x^{2}+2)^{\frac{1}{3}}$

Let $u=-x^{2}+2$, then $y = u^{\frac{1}{3}}$.
$\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$ and $\frac{du}{dx}=-2x$.
$\frac{dy}{dx}=\frac{1}{3}(-x^{2}+2)^{-\frac{2}{3}}\cdot(-2x)=\frac{-2x}{3\sqrt[3]{(-x^{2}+2)^{2}}}$.

1f. For $y=\sec^{2}(5x)$

Let $u = \sec(5x)$, then $y = u^{2}$.
For $u=\sec(5x)$, let $t = 5x$, $u=\sec(t)$, $\frac{du}{dt}=\sec(t)\tan(t)$ and $\frac{dt}{dx}=5$, so $\frac{du}{dx}=5\sec(5x)\tan(5x)$.
$\frac{dy}{du}=2u$, so $\frac{dy}{dx}=2\sec(5x)\cdot5\sec(5x)\tan(5x)=10\sec^{2}(5x)\tan(5x)$.

2. For the equation $x + y^{3}-y=1$

Differentiate implicitly with respect to $x$:
$1 + 3y^{2}\frac{dy}{dx}-\frac{dy}{dx}=0$.
$(3y^{2}-1)\frac{dy}{dx}=-1$.
$\frac{dy}{dx}=\frac{-1}{3y^{2}-1}$.
When $x = 1$, $1+y^{3}-y = 1$, so $y^{3}-y=0$, $y(y - 1)(y + 1)=0$, $y=0,1,-1$.
When $y = 0$, $\frac{dy}{dx}=1$, and the tangent - line equation is $y-0=1\cdot(x - 1)$, i.e., $y=x - 1$.
When $y = 1$, $\frac{dy}{dx}=-\frac{1}{2}$, and the tangent - line equation is $y - 1=-\frac{1}{2}(x - 1)$, i.e., $y=-\frac{1}{2}x+\frac{3}{2}$.
When $y=-1$, $\frac{dy}{dx}=-\frac{1}{2}$, and the tangent - line equation is $y + 1=-\frac{1}{2}(x - 1)$, i.e., $y=-\frac{1}{2}x-\frac{1}{2}$.

3. Let $h$ be the height of the elevator above the ground, the horizontal distance from the observer to the elevator shaft is $x = 20$ m and the observer is $20$ m above the ground.

The angle of elevation $\theta$ satisfies $\tan\theta=\frac{h - 20}{20}$.
Differentiate both sides with respect to time $t$:
$\sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{20}\frac{dh}{dt}$.
We know that $\frac{dh}{dt}=5$ m/s.
When $h = 10$ m, $\tan\theta=\frac{10 - 20}{20}=-\frac{1}{2}$, $\sec^{2}\theta=1+\tan^{2}\theta=1+\frac{1}{4}=\frac{5}{4}$.
$\frac{5}{4}\frac{d\theta}{dt}=\frac{1}{20}\times5$, $\frac{d\theta}{dt}=\frac{1}{5}$ rad/s.
When $h = 40$ m, $\tan\theta=\frac{40 - 20}{20}=1$, $\sec^{2}\theta=1 + 1=2$.
$2\frac{d\theta}{dt}=\frac{1}{20}\times5$, $\frac{d\theta}{dt}=\frac{1}{8}$ rad/s.

Answer:

1a. $\frac{dy}{dx}=15x^{2}(x^{3}+1)^{4}$
1b. $\frac{dy}{dx}=\frac{9x^{2}}{2\sqrt{3x^{3}-4}}$
1c. $\frac{dy}{dx}=4x^{3}\cos(x^{4})$
1d. $\frac{dy}{dx}=2x(2x^{5}+9)^{3}(22x^{5}+60x^{3}+9)$
1e. $\frac{dy}{dx}=\frac{-2x}{3\sqrt[3]{(-x^{2}+2)^{2}}}$
1f. $\frac{dy}{dx}=10\sec^{2}(5x)\tan(5x)$

1. $y=x - 1$, $y=-\frac{1}{2}x+\frac{3}{2}$, $y=-\frac{1}{2}x-\frac{1}{2}$
2. When $h = 10$ m, $\frac{d\theta}{dt}=\frac{1}{5}$ rad/s; when $h = 40$ m, $\frac{d\theta}{dt}=\frac{1}{8}$ rad/s