Solve the problem.
-If $a$ and $b$ are the lengths of the legs of a right triangle and $c$ is the length of the hypotenuse, $c ^ { 2 } = a ^ { 2 } + b ^ { 2 }$. How is $\mathrm { dc } / \mathrm { d }$ t related to $\mathrm { da } / \mathrm { d }$ t and $\mathrm { db } / \mathrm { dt }$ ?
A) $\frac { \mathrm { dc } } { \mathrm { dt } } = \mathrm { a } \frac { \mathrm { da } } { \mathrm { dt } } + \mathrm { b } \frac { \mathrm { db } } { \mathrm { dt } }$
B) $\frac { \mathrm { dc } } { \mathrm { dt } } = \mathrm { a } ^ { 2 } \frac { \mathrm { da } } { \mathrm { dt } } + \mathrm { b } ^ { 2 } \frac { \mathrm { db } } { \mathrm { dt } }$
C) $\frac { \mathrm { dc } } { \mathrm { dt } } = \frac { 1 } { \mathrm { c } } \left( \mathrm { a } \frac { \mathrm { da } } { \mathrm { dt } } + \mathrm { b } \frac { \mathrm { db } } { \mathrm { dt } } \right)$
D) $\frac { d c } { d t } = 2 a \frac { d a } { d t } + 2 b \frac { d b } { d t }$

Question

Quizplus · Accepted Answer

Taking the derivative of both sides with respect to time, we get

$$2c \frac{dc}{dt} = 2a \frac{da}{dt} + 2b \frac{db}{dt}$$

Dividing both sides by $2c$, we get

$$\frac{dc}{dt} = \frac{a}{c} \frac{da}{dt} + \frac{b}{c} \frac{db}{dt}$$

Using the Pythagorean Theorem $c^2 = a^2 + b^2$, we can simplify the expression to

$$\frac{dc}{dt} = \frac{1}{c}(a\frac{da}{dt} + b\frac{db}{dt})$$

Therefore, the answer is (C).

Solve the Problem aaa And bbb Are the Lengths of the Legs of a Right Triangle

Solve the Problem $a$ And $b$ Are the Lengths of the Legs of a Right Triangle