Problem Set:Problem 109 (1 point) Let g(s)=e^s(s^4+3) a) Find the equation of the tangent line to g at (0,3) y= b) Find the equation of the normal line to g at (0,3) y=

To find the equation of the tangent line to the function \( g(s) = e^s(s^4 + 3) \) at the point \((0, 3)\), we need to follow these steps:<br /><br />1. **Find the derivative \( g'(s) \) to get the slope of the tangent line.**<br />2. **Evaluate the derivative at \( s = 0 \) to find the slope of the tangent line at the point \((0, 3)\).**<br />3. **Use the point-slope form of the equation of a line to write the equation of the tangent line.**<br /><br />Let's go through these steps:<br /><br />### Step 1: Find the derivative \( g'(s) \)<br /><br />Given \( g(s) = e^s(s^4 + 3) \), we use the product rule to differentiate:<br /><br />\[ g'(s) = \frac{d}{ds} [e^s(s^4 + 3)] \]<br /><br />The product rule states that if \( u(s) \) and \( v(s) \) are functions of \( s \), then:<br /><br />\[ \frac{d}{ds} [u(s)v(s)] = u'(s)v(s) + u(s)v'(s) \]<br /><br />Here, let \( u(s) = e^s \) and \( v(s) = s^4 + 3 \).<br /><br />\[ u'(s) = e^s \]<br />\[ v'(s) = 4s^3 \]<br /><br />Now apply the product rule:<br /><br />\[ g'(s) = e^s(s^4 + 3) + e^s(4s^3) \]<br />\[ g'(s) = e^s(s^4 + 3 + 4s^3) \]<br />\[ g'(s) = e^s(s^4 + 4s^3 + 3) \]<br /><br />### Step 2: Evaluate the derivative at \( s = 0 \)<br /><br />\[ g'(0) = e^0(0^4 + 4 \cdot 0^3 + 3) \]<br />\[ g'(0) = 1 \cdot 3 \]<br />\[ g'(0) = 3 \]<br /><br />So, the slope of the tangent line at \( (0, 3) \) is 3.<br /><br />### Step 3: Write the equation of the tangent line<br /><br />The point-slope form of the equation of a line is:<br /><br />\[ y - y_1 = m(x - x_1) \]<br /><br />Here, \( (x_1, y_1) = (0, 3) \) and \( m = 3 \).<br /><br />\[ y - 3 = 3(x - 0) \]<br />\[ y - 3 = 3x \]<br />\[ y = 3x + 3 \]<br /><br />So, the equation of the tangent line is:<br /><br />\[ y = 3x + 3 \]<br /><br />### Step 4: Find the equation of the normal line<br /><br />The slope of the normal line is the negative reciprocal of the slope of the tangent line. The slope of the tangent line is 3, so the slope of the normal line is:<br /><br />\[ -\frac{1}{3} \]<br /><br />Using the point-slope form again with \( (x_1, y_1) = (0, 3) \) and \( m = -\frac{1}{3} \):<br /><br />\[ y - 3 = -\frac{1}{3}(x - 0) \]<br />\[ y - 3 = -\frac{1}{3}x \]<br />\[ y = -\frac{1}{3}x + 3 \]<br /><br />So, the equation of the normal line is:<br /><br />\[ y = -\frac{1}{3}x + 3 \]<br /><br />### Final Answers<br /><br />a) The equation of the tangent line to \( g \) at \( (0, 3) \) is:<br /><br />\[ y = 3x + 3 \]<br /><br />b) The equation of the normal line to \( g \) at \( (0, 3) \) is:<br /><br />\[ y = -\frac{1}{3}x + 3 \]