Question

Find y=mx+b of 3xcosx at (π,−3π).

a) y=−3πx
b) y=3πx
c) y=−3πx−3π
d) y=3πx−3π

Answer 1

After differentiating y = 3xcos(x) to find the slope, the equation of the tangent line that passes through the point (π,−3π) is y = -3πx, which corresponds to option (a).

Step-by-step explanation:

The student is asking for the equation of a line in the form y=mx+b where m is the slope and b is the y-intercept, given the point (π,−3π) which lies on the tangent to the curve y = 3xcos(x). To find this equation, we can use the derivative of 3xcos(x) to find the slope at x=π, which is the value of m, and then use the point to solve for b.

First, we differentiate y = 3xcos(x) with respect to x to find the slope at x=π:

dy/dx = 3cos(x) - 3xsin(x)

At x=π, the cosine of π is -1 and the sine of π is 0, so:

dy/dx = 3(-1) - 3(π)(0) = -3

The slope of the tangent line at (π,−3π) is -3. Hence, m = -3.

Now, we use the point (π,−3π) to find b:

y = mx + b

-3π = -3(π) + b

b = 0

The equation of the line is therefore y = -3πx, which corresponds to option (a) y=−3πx.