Sandra Allison

Answered

2021-12-22

Suppose

(a) if

(b) if

Answer & Explanation

SlabydouluS62

Expert

2021-12-23Added 52 answers

Given that $y=\sqrt{2x+1}$

x and y are functions of t.

a)$\frac{dx}{dt}=3$

To find$\frac{dy}{dt}-3$ when $x=4$

$y=\sqrt{2x+1}$

$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$

$=\frac{1}{2\sqrt{2x+1}}\cdot 2\cdot \frac{dx}{dt}$

$=\frac{1}{\sqrt{2x+1}}\cdot 3$

$=\frac{3}{\sqrt{2\cdot 4+1}}$ at $x=4$

$=\frac{3}{\sqrt{9}}$

$=\frac{3}{3}$

$=1$

So$\frac{dy}{dt}=1$ at $x=4$

x and y are functions of t.

a)

To find

So

Ronnie Schechter

Expert

2021-12-24Added 27 answers

b) Given that $\frac{dy}{dt}=5$ . To find $\frac{dx}{dt}$ when $x=12$

$y=\sqrt{2x+1}$

${y}^{2}=2x+1$ (On squaring both sides)

$x=\frac{1}{2}({y}^{2}-1)$

$\frac{dx}{dt}=\frac{1}{2}\left(2y\right)=y$

$\frac{dx}{dt}=\frac{dx}{dy}\cdot \frac{dy}{dt}$

$=y\cdot \frac{dy}{dt}$

$=5\cdot 5$ (when $x=12$ we get $y=\sqrt{2\cdot 12+1}=\sqrt{25}=5$ )

$=25$

So$\frac{dx}{dt}=25$

So

user_27qwe

Expert

2021-12-30Added 230 answers

Very helpful! I used it to solve a similar problem.

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