Suppose y=2x+1 where x and y are functions of t. (a) if dxdt=3,finddydtwhen x=4. (b) if dydt=5, find dxdt when x=12.

Given that y=2x+1 x and y are functions of t. a) dxdt=3 To find dydt−3 when x=4 y=2x+1 dydt=dydx⋅dxdt =122x+1⋅2⋅dxdt =12x+1⋅3 =32⋅4+1 at x=4 =39 =33 =1 So dydt=1 at x=4

b) Given that dydt=5. To find dxdt when x=12 y=2x+1 y2=2x+1 (On squaring both sides) x=12(y2−1) dxdt=12(2y)=y dxdt=dxdy⋅dydt =y⋅dydt =5⋅5 (when x=12 we get y=2⋅12+1=25=5) =25 So dxdt=25

Given that y=2x+1 x and y are functions of t. a) dxdt=3 To find dydt−3 when x=4 y=2x+1 dydt=dydx⋅dxdt =122x+1⋅2⋅dxdt =12x+1⋅3 =32⋅4+1 at x=4 =39 =33 =1 So dydt=1 at x=4

b) Given that dydt=5. To find dxdt when x=12 y=2x+1 y2=2x+1 (On squaring both sides) x=12(y2−1) dxdt=12(2y)=y dxdt=dxdy⋅dydt =y⋅dydt =5⋅5 (when x=12 we get y=2⋅12+1=25=5) =25 So dxdt=25

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Sandra Allison

Answered question

2021-12-22

Suppose $y = \sqrt{2 x + 1}$ where x and y are functions of t.
(a) if $\frac{d x}{d t} = 3, f i n d \frac{d y}{d t} w h e n x = 4$ .
(b) if $\frac{d y}{d t} = 5$ , find $\frac{d x}{d t}$ when $x = 12 .$

Answer & Explanation

SlabydouluS62

Skilled2021-12-23Added 52 answers

Given that

y = \sqrt{2 x + 1}

x and y are functions of t.
a)

\frac{d x}{d t} = 3

To find

\frac{d y}{d t} - 3

when

x = 4

y = \sqrt{2 x + 1}

\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}

= \frac{1}{2 \sqrt{2 x + 1}} \cdot 2 \cdot \frac{d x}{d t}

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

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

x = 4

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

= \frac{3}{3}

= 1

\frac{d y}{d t} = 1

x = 4

Ronnie Schechter

Beginner2021-12-24Added 27 answers

b) Given that

\frac{d y}{d t} = 5

. To find

\frac{d x}{d t}

when

x = 12

y = \sqrt{2 x + 1}

y^{2} = 2 x + 1

(On squaring both sides)

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

\frac{d x}{d t} = \frac{1}{2} (2 y) = y

\frac{d x}{d t} = \frac{d x}{d y} \cdot \frac{d y}{d t}

= y \cdot \frac{d y}{d t}

= 5 \cdot 5

(when

x = 12

we get

y = \sqrt{2 \cdot 12 + 1} = \sqrt{25} = 5

)

= 25

\frac{d x}{d t} = 25

user_27qwe

Skilled2021-12-30Added 375 answers

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

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