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How do you find the instantaneous rate of change for the volume of a growing spherical cell given by #v = (4/3) (pi) r^3# when r is 5?

Answer 1

Firs evaluate the derivative of your volume (with respect to #r#: #V'(r)=(dV(r))/(dr)=(4/3)pi*3r^2# Now substitute #r=5# in it: #V'(5)=(4/3)pi*3*25=100pi#

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Answer 2

Elijah Abram

To find the instantaneous rate of change for the volume of a growing spherical cell at ( r = 5 ), you need to take the derivative of the volume function ( v = \frac{4}{3} \pi r^3 ) with respect to ( r ) and then evaluate it at ( r = 5 ). The derivative of ( v ) with respect to ( r ) is ( \frac{dv}{dr} = 4 \pi r^2 ). Evaluating this at ( r = 5 ), we get ( \frac{dv}{dr} = 4 \pi (5)^2 = 100 \pi ) cubic units per unit length. Therefore, the instantaneous rate of change for the volume of the spherical cell when ( r = 5 ) is ( 100 \pi ) cubic units per unit length.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.