Question

Find equations for the lines that are tangent and normal to the graph of y=x^{2} \sin x\( at \)x=3

Short answer

The equations of the tangent and normal lines are respectively given by:\[ y -3^{2} \sin 3 = m_{t} (x - 3)\] \[ y -3^{2} \sin 3 = m_{n} (x - 3)\]where \(m_{t}\) is the slope of the tangent line and \(m_{n}\) is the slope of the normal line. These slopes can be calculated by substituting \(x = 3\) in the derivative of the function and its reciprocal respectively.

Step-by-step solution

Find the derivative of the function

The function is \(y = x^{2} \sin x\). Recall that the derivative of a product of two functions is obtained using the product rule, i.e. \((uv)' = u'v + uv'\), where \(u = x^{2}\) and \(v = \sin x\). Thus it follows that, \[y' = (x^{2})' \sin x + x^{2} (\sin x)'\] Hence \[y' = 2x \sin x + x^{2} \cos x\]

Calculate the slope of the tangent line at x = 3

Substitute \(x = 3\) into the derivative equation obtained in step 1. So the slope of the tangent line, \(m_{t}\), is given by: \[m_{t} = 2*3 \sin 3 + 3^{2} \cos 3\]

Find the equation of the tangent line

The equation of a line in point-slope form (y - y1 = m(x - x1)) requires a point and slope. We have the slope from the previous step, \(m_{t}\). The point on the line can be obtained by evaluating function y at x = 3 - giving \((x_1, y_1)\) = \((3, 3^{2} \sin 3)\). Plugging these values into the point-slope form equation gives the equation of the tangent line.

Calculate the slope of the normal line

The slope of the normal line, \(m_{n}\), is the negative reciprocal of the slope of the tangent line. Therefore, \[m_{n} = -1/m_{t}\]

Find the equation of the normal line

Use the slope from Step 4 and the same point from Step 3, \((x_1, y_1)\), to find the equation of the normal line using the point-slope form equation from Step 3.