Math / Calculus

Question3 Variables $x$ and $y$ are connected by the equation $y=(x-3)(x+5)^{3}$ . Given that $x$ increases at a rate of 0.2 uats per second, find the rate of change of $y$ when $x=-4$ .

Studdy Solution

STEP 1

1. The function $y = (x-3)(x+5)^3$ describes the relationship between $x$ and $y$ .
2. We are given that $\frac{dx}{dt} = 0.2$ units per second.
3. We need to find $\frac{dy}{dt}$ when $x = -4$ .

STEP 2

1. Differentiate the function $y = (x-3)(x+5)^3$ with respect to $x$ .
2. Use the chain rule to find $\frac{dy}{dt}$ .
3. Substitute the given values to find the rate of change of $y$ .

STEP 3

Differentiate $y = (x-3)(x+5)^3$ with respect to $x$ . This requires the use of the product rule and the chain rule.
Let $u = x-3$ and $v = (x+5)^3$ . Then $y = uv$ .
The derivative of $y$ with respect to $x$ is given by:
$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$
First, find $\frac{du}{dx}$ and $\frac{dv}{dx}$ :
$\frac{du}{dx} = 1$
$v = (x+5)^3 \Rightarrow \frac{dv}{dx} = 3(x+5)^2 \cdot 1 = 3(x+5)^2$
Now substitute back:
$\frac{dy}{dx} = (x-3) \cdot 3(x+5)^2 + (x+5)^3 \cdot 1$
$\frac{dy}{dx} = 3(x-3)(x+5)^2 + (x+5)^3$

STEP 4

Use the chain rule to find $\frac{dy}{dt}$ :
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

STEP 5

Substitute $x = -4$ and $\frac{dx}{dt} = 0.2$ into the expression for $\frac{dy}{dt}$ .
First, calculate $\frac{dy}{dx}$ at $x = -4$ :
$\frac{dy}{dx} = 3(-4-3)(-4+5)^2 + (-4+5)^3$
$= 3(-7)(1)^2 + (1)^3$
$= 3(-7)(1) + 1$
$= -21 + 1$
$= -20$
Now, calculate $\frac{dy}{dt}$ :
$\frac{dy}{dt} = -20 \cdot 0.2$
$= -4$
The rate of change of $y$ when $x = -4$ is:
$\boxed{-4}$ units per second.

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms