Recall that the volume of a sphere of radius r is V(r) =4\, \pi\, r^3 /3.
Find L, the linearisation of V(r) at r=50
To find the linearization of V(r) at r = 50, we need to find the equation of the tangent line to the curve at that point. The equation of a tangent line can be written in the form y = mx + b, where m represents the slope of the line and b represents the y-intercept.
First, let's start by finding the derivative of V(r) with respect to r:
V'(r) = dV(r)/dr = 4π (3r^2)/3 = 4π r^2
Next, we need to evaluate the derivative at r = 50 to find the slope of the tangent line at that point:
m = V'(50) = 4π (50)^2 = 4π (2500) = 10000π
So far, the equation of the tangent line is y = 10000πx + b. To find the value of b, we need to substitute the values of x and y corresponding to the point (r, V(r)) = (50, V(50)).
V(50) = (4π (50)^3) /3 = (4π (125000) )/3 = (500000π)/3
Now, we can substitute the values into the equation of the tangent line:
500000π/3 = 10000π (50) + b
Simplifying the equation:
500000π/3 = 500000π + b
Now, we can isolate b:
b = 500000π/3 - 500000π
Simplifying further:
b = 500000π (1/3 - 1)
b = 500000π (1/3 - 3/3)
b = 500000π (-2/3)
b = -1000000π/3
Finally, we have the equation of the tangent line (linearization):
L(x) = 10000πx - 1000000π/3
Therefore, the linearization of V(r) at r = 50 is L(x) = 10000πx - 1000000π/3.
v = 4π/3 r^3
v' = 4π r^2
at r=50, v = 500π/3 and v' = 100π
So, the tangent line at r=50 is
v - 500π/3 = 100π(x-50)