Find $\cfrac{dy}{dt}$ for the given values of $x$.

Question

Question:

A point is moving along the graph of the given function at the rate $\cfrac{dx}{dt}$. Find $\cfrac{dy}{dt}$ for the given values of $x$.

$y = 4x^2 + 1$; $\cfrac {dx}{dt} = 4$ centimeters per second.

a. $x=1$

I'm a bit stuck on this problem. I don't understand what I'm supposed to plug into. Any help would be appreciated!

Answer 1

You are supposed to use $\frac {dy}{dt}=\frac {dy}{dx}\cdot \frac {dx}{dt}$

Answer 2

We have

\begin{align} y&=4x^2+1 \\ \frac{dy}{dx} & = 8x \end{align}

By product rule,

\begin{align} \frac{dy}{dt} & = \frac{dy}{dx} \cdot \frac{dx}{dt} \\ & = 8x \cdot4 \\ & = 32 x \end{align}

Now just plug in $x=1$:

$$32x = 32 \cdot 1 = 32$$

Hence the rate of change of $y$ wrt $t$ is $32$.

Answer 3

Use the chain rule...

$\frac{dy}{dx}=8x$. Then $\frac {dy}{dt}=\frac {dy}{dx}\cdot \frac {dx}{dt}=8x\cdot4=32x=32\cdot1=32$ at $x=1$...

$32$ centimeters per second.

Answer 4

$y(t)= 4x^2(t) +1.$

Differentiate both sides with respect to $t:$

$\dfrac{dy}{dt}= \dfrac{d(4x^2+1)}{dx}\dfrac{dx}{dt}.$

$\dfrac{dy}{dt}= (8x)\dfrac{dx}{dt}.$

At $x=1$, $\dfrac{dx}{dt}=4$, find $\dfrac{dy}{dt}.$

What is left to do?