homework5: problem 19 (1 point)\nlet $g(s)=e^{s}(s^{2}+7)$.\na. find the equation of the tangent line to $g$…

Explanation:

Step1: Find the derivative of $g(s)$ using product - rule

The product - rule states that if $y = u(s)v(s)$, then $y^\prime=u^\prime(s)v(s)+u(s)v^\prime(s)$. Here, $u(s)=e^{s}$ and $v(s)=s^{2}+7$. The derivative of $u(s)$ is $u^\prime(s)=e^{s}$, and the derivative of $v(s)$ is $v^\prime(s) = 2s$. So, $g^\prime(s)=e^{s}(s^{2}+7)+e^{s}(2s)=e^{s}(s^{2}+2s + 7)$.

Step2: Evaluate the derivative at $s = 0$ to get the slope of the tangent line

Substitute $s = 0$ into $g^\prime(s)$. $g^\prime(0)=e^{0}(0^{2}+2\times0 + 7)=7$.

Step3: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$. Here, $(x_{1},y_{1})=(0,7)$ and $m = 7$. So, $y-7=7(x - 0)$, which simplifies to $y=7x + 7$.

Step4: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line $m_{t}=7$, the slope of the normal line $m_{n}=-\frac{1}{7}$.

Step5: Find the equation of the normal line

Using the point - slope form with $(x_{1},y_{1})=(0,7)$ and $m =-\frac{1}{7}$, we have $y - 7=-\frac{1}{7}(x - 0)$, which simplifies to $y=-\frac{1}{7}x+7$.

Answer:

a. $y = 7x+7$ b. $y=-\frac{1}{7}x + 7$