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Assume that [tex]$x$[/tex] and [tex]$y$[/tex] are both differentiable functions of [tex][tex]$t$[/tex][/tex]. Find the required values of [tex]$\frac{dy}{dt}$[/tex] and [tex]$\frac{dx}{dt}$[/tex].

[tex]y = \sqrt{x}[/tex]

(a) Find [tex][tex]$\frac{dy}{dt}$[/tex][/tex], given [tex]$x = 16$[/tex] and [tex]$\frac{dx}{dt} = 2$[/tex].

[tex]\frac{dy}{dt} = \square[/tex]

(b) Find [tex][tex]$\frac{dx}{dt}$[/tex][/tex], given [tex]$x = 64$[/tex] and [tex]$\frac{dy}{dt} = 5$[/tex].

[tex]\frac{dx}{dt} = \square[/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's solve the problem step-by-step for each part.

### (a) Finding [tex]\(\frac{dy}{dt}\)[/tex] given [tex]\(x = 16\)[/tex] and [tex]\(\frac{dx}{dt} = 2\)[/tex]

Given the relationship:
[tex]\[ y = \sqrt{x} \][/tex]

To find [tex]\(\frac{dy}{dt}\)[/tex], we use the chain rule. The chain rule states:
[tex]\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \][/tex]

First, we need to find [tex]\(\frac{dy}{dx}\)[/tex]. Differentiating [tex]\(y = \sqrt{x}\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ y = x^{1/2} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} \][/tex]

Now, substituting [tex]\(x = 16\)[/tex]:
[tex]\[ \frac{dy}{dx} \Bigg|_{x = 16} = \frac{1}{2} \cdot \frac{1}{\sqrt{16}} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \][/tex]

We are given [tex]\(\frac{dx}{dt} = 2\)[/tex]. Therefore, we can find [tex]\(\frac{dy}{dt}\)[/tex]:
[tex]\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{1}{8} \cdot 2 = \frac{1}{4} \][/tex]
[tex]\[ \frac{dy}{dt} = 0.25 \][/tex]

So, [tex]\(\boxed{0.25}\)[/tex] is the value of [tex]\(\frac{dy}{dt}\)[/tex] for part (a).

### (b) Finding [tex]\(\frac{dx}{dt}\)[/tex] given [tex]\(x = 64\)[/tex] and [tex]\(\frac{dy}{dt} = 5\)[/tex]

Given the relationship:
[tex]\[ y = \sqrt{x} \][/tex]

Using the chain rule again:
[tex]\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \][/tex]

We need to find [tex]\(\frac{dx}{dt}\)[/tex]. First, focusing on [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} \][/tex]

Now, substituting [tex]\(x = 64\)[/tex]:
[tex]\[ \frac{dy}{dx} \Bigg|_{x = 64} = \frac{1}{2} \cdot \frac{1}{\sqrt{64}} = \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16} \][/tex]

Given [tex]\(\frac{dy}{dt} = 5\)[/tex], we solve for [tex]\(\frac{dx}{dt}\)[/tex]:
[tex]\[ \frac{dy}{dt} = \frac{1}{16} \cdot \frac{dx}{dt} \][/tex]
[tex]\[ 5 = \frac{1}{16} \cdot \frac{dx}{dt} \][/tex]
[tex]\[ \frac{dx}{dt} = 5 \cdot 16 = 80 \][/tex]

So, [tex]\(\boxed{80}\)[/tex] is the value of [tex]\(\frac{dx}{dt}\)[/tex] for part (b).

In summary:
- For part (a), [tex]\(\frac{dy}{dt} = 0.25\)[/tex]
- For part (b), [tex]\(\frac{dx}{dt} = 80\)[/tex]
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