Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

One leg of a right triangle is decreasing at a rate of 5 kilometers per hour and the other leg of the triangle is increasing at a rate of 14 kilometers per hour.
At a certain instant, the decreasing leg is 3 kilometers and the increasing leg is 9 kilometers.
What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)?
Choose 1 answer:
(A) 1.5
(B) 111
(C) -1.5
(D) -111

One leg of a right triangle is decreasing at a rate of $5$ kilometers per hour and the other leg of the triangle is increasing at a rate of $14$ kilometers per hour. $\newline$ At a certain instant, the decreasing leg is $3$ kilometers and the increasing leg is $9$ kilometers. $\newline$ What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ . $5$ $\newline$ (B) $111$ $\newline$ (C) $-1$ . $5$ $\newline$ (D) $-111$

View step-by-step help

View step-by-step help

Find instantaneous rates of change

Full solution

Q. One leg of a right triangle is decreasing at a rate of

5

kilometers per hour and the other leg of the triangle is increasing at a rate of

14

kilometers per hour.

\newline

At a certain instant, the decreasing leg is

3

kilometers and the increasing leg is

9

kilometers.

\newline

What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)?

\newline

Choose

1

answer:

\newline

(A)

1

.

5

\newline

(B)

111

\newline

(C)

-1

.

5

\newline

(D)

-111

Area Formula Explanation: The area of a right triangle is given by the formula $A = (\frac{1}{2}) \times \text{base} \times \text{height}$ . Here, one leg is the base and the other is the height.
Legs Notation: Let's denote the decreasing leg as ' $b$ ' and the increasing leg as ' $h$ '. The rate of change of ' $b$ ' is $-5$ km/h and the rate of ' $h$ ' is $14$ km/h.
Given Rates: At the instant in question, $b = 3 \, \text{km}$ and $h = 9 \, \text{km}$ . The rate of change of the area $A$ with respect to time $t$ can be found by differentiating the area formula with respect to $t$ .
Instant Values: $\frac{dA}{dt} = \frac{1}{2} \cdot \left(\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}\right)$ . Now we plug in the values: $\frac{db}{dt} = -5 \, \text{km/h}$ , $\frac{dh}{dt} = 14 \, \text{km/h}$ , $b = 3 \, \text{km}$ , and $h = 9 \, \text{km}$ .
Differentiation: $\frac{dA}{dt} = \frac{1}{2} \times (-5 \times 9 + 3 \times 14)$ .
Substitute Values: $\frac{dA}{dt} = \frac{1}{2} \times (-45 + 42)$ .
Calculate: $\frac{dA}{dt} = \frac{1}{2} \times (-3)$ .
Final Result: $\frac{dA}{dt} = -1.5$ square kilometers per hour. So, the correct answer is (C) $-1.5$ .

More problems from Find instantaneous rates of change

Question

The graphs of the functions

f(x)=\sin (x)

g(x)=\frac{1}{2}

2

points on the interval

0<x<\pi

\newline

What is the area of the region bound by the graphs of

f(x)

g(x)

between those points of intersection ?

\newline

1

\newline

\frac{\pi}{3}

\newline

\frac{\pi}{2}

\newline

2-\frac{\pi}{2}

\newline

\sqrt{3}-\frac{\pi}{3}

Posted 3 months ago

Question

What is the area of the region between the graphs of

f(x)=x^{2}+12 x

g(x)=3 x^{2}+10

x=1

x=4

\newline

1

\newline

77

\newline

\frac{64}{3}

\newline

18

\newline

45

Posted 3 months ago

Question

What is the area of the region between the graphs of

f(x)=\sqrt{x+1}

g(x)=2 x-4

x=0

x=3

\newline

1

\newline

\frac{23}{3}

\newline

\frac{5}{3}

\newline

\frac{14}{3}

\newline

-3

Posted 3 months ago

Question

Consider the curve given by the equation

3 x^{2}+y^{4}+6 x=253

. It can be shown that

\frac{d y}{d x}=\frac{-6(x+1)}{4 y^{3}}.

\newline

Write the equation of the horizontal line that is tangent to the curve and is above the

x

Posted 3 months ago

Question

The learning rate for new skills is proportional to the difference between the maximum potential for learning that skill,

M

, and the amount of the skill already learned,

L

\newline

Which equation describes this relationship?

\newline

1

\newline

L(t)=\frac{k}{(M-L)}

\newline

L(t)=k(M-L)

\newline

\frac{d L}{d t}=k(M-L)

\newline

\frac{d L}{d t}=\frac{k}{(M-L)}

Posted 3 months ago

Question

What is the value of

\frac{d}{d x}\left(\frac{1}{x}\right)

x=6

Posted 3 months ago

Question

What is the value of

\frac{d}{d x}(\sqrt{x})

x=9

Posted 3 months ago

Question

What is the value of

\frac{d}{d x}\left(x^{-3}\right)

x=3

Posted 3 months ago

Question

The exponential function

f

is graphed in the

x y

x

1, y

increases by a factor of

3

. Which of the following could be

f

\newline

1

\newline

f(x)=\left(\frac{1}{3}\right)^{x}

\newline

f(x)=\left(\frac{1}{3}\right)^{x}+3

\newline

f(x)=3^{x}+2

\newline

f(x)=2(3)^{x}

Posted 3 months ago

Question

What are the critical points for the plane curve defined by the equations

\newline

x(t)=\cot t

y(t)=\sin t

\newline

0 < t < \pi

? Write your answer as a list of values of

\newline

t

, separated by commas. For example, if you found

\newline

t=1

\newline

t=2

, you would enter

1,2

\newline

Provide your answer below:

Posted 2 months ago