4. the altitude of a triangle is increasing at a rate of 3 cm/sec while the area of the triangle is…

Explanation:

Step1: Recall area formula

The area formula of a triangle is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the altitude.

Step2: Differentiate with respect to time

Differentiate both sides of the equation $A = \frac{1}{2}bh$ with respect to time $t$ using the product - rule. The product - rule states that if $y = uv$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Here, $u = b$ and $v = h$, so $\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})$.

Step3: Substitute given values

We are given that $\frac{dh}{dt}=3$ cm/sec, $\frac{db}{dt}=1$ cm/sec, $h = 10$ cm and $A = 70$ cm². First, from $A=\frac{1}{2}bh$, we can find $b$. Since $70=\frac{1}{2}b\times10$, then $b = 14$ cm. Substitute $b = 14$ cm, $h = 10$ cm, $\frac{dh}{dt}=3$ cm/sec and $\frac{db}{dt}=1$ cm/sec into $\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})$. $\frac{dA}{dt}=\frac{1}{2}(14\times3 + 10\times1)$.

Step4: Calculate the result

$\frac{dA}{dt}=\frac{1}{2}(42 + 10)=\frac{1}{2}\times52=26$ cm²/sec.

Answer:

26 cm²/sec