Question

Find an equation of the line that is both tangent to the curve y=x ⁴+1 and parallel to the line 32x−y=15. 63. Find equations for two lines that are both tangent to the curve y=x ³ −3x ² +3x−3 and parallel to the line 3x−y=15

Answer 1

To find an equation of the line that is both tangent to the curve y = x⁴ + 1 and parallel to the line 32x - y = 15, we need to find the slope of the curve at the point of tangency. The equation of the tangent line is y = 32x + (32a + 1).

Step-by-step explanation:

To find an equation of the line that is both tangent to the curve y = x⁴ + 1 and parallel to the line 32x - y = 15, we need to find the slope of the curve at the point of tangency. Taking the derivative of the curve, we get y' = 4x³. Since the line is parallel to 32x - y = 15, it has the same slope of 32. Therefore, the equation of the tangent line is y = 32x + b, where b is the y-intercept.

To find b, we substitute the x-coordinate of the point of tangency, let's call it (a, b), into the equation y = 32x + b. Substituting (a, b), we get b = 32a + b. Solving for b, we get b = 32a + 1.

So, the equation of the line tangent to y = x⁴ + 1 and parallel to 32x - y = 15 is y = 32x + (32a + 1).

Answer 2

The equation of the line tangent to the curve y=x⁴+1 and parallel to the line 32x−y=15 is y=x⁴ −32x+32.

For the curve y=x³−3x² +3x−3, two lines that are tangent and parallel to 3x−y=15 have the equations y=x³+3x−7 and y=x³−3x² +3x−11.

Step-by-step explanation:

To find the equation of the tangent line to the curve y=x⁴+1 and parallel to 32x−y=15, we need the slopes to be equal. The slope of the curve at any point is given by the derivative y′. The slope of the line 32x−y=15 is found by rearranging it to y=32x−15. Equating these slopes, we get 4x³=32, solving which yields x=2. Substituting x=2 back into the curve equation gives y=17. Thus, the tangent line equation is y=x⁴−32x+32.

Similarly, for the curve y=x³ −3x² +3x−3, we find its derivative y′. Setting y′=3x² −6x+3 equal to the slope of the line 3x−y=15, we solve for x and get x=2. Substituting x=2 into the curve equation gives y=−7. Therefore, the equation of the tangent line is y=x³−3x² +3x−7. Similarly, another tangent line is found by substituting x=2 into the curve equation y=x³−3x²+3x−11.

By calculating the slopes and using the point of tangency, we determine the equations for the required tangent lines.