Find the gradient of the tangent to the curve [tex]w=\frac{v^3-7 v+2}{(v+2)(v^2-v+3)}[/tex] at the point [tex]v=\frac{2}{3}[/tex]
Answer (1)
Find the derivative d v d w of the given function w = ( v + 2 ) ( v 2 − v + 3 ) v 3 − 7 v + 2 using the quotient rule.
Evaluate the derivative at the point v = 3 2 .
Simplify the expression to find the gradient of the tangent.
The gradient of the tangent to the curve at v = 3 2 is − 5000 3033 .
Explanation
Problem Analysis We are given the curve w = ( v + 2 ) ( v 2 − v + 3 ) v 3 − 7 v + 2 and we want to find the gradient of the tangent to this curve at the point v = 3 2 . The gradient of the tangent is given by the derivative d v d w evaluated at v = 3 2 .
Finding the Derivative First, we need to find the derivative of w with respect to v . Let's denote the numerator as f ( v ) = v 3 − 7 v + 2 and the denominator as g ( v ) = ( v + 2 ) ( v 2 − v + 3 ) = v 3 + v + 6 . Then w = g ( v ) f ( v ) . We can use the quotient rule to find the derivative: d v d w = [ g ( v ) ] 2 f ′ ( v ) g ( v ) − f ( v ) g ′ ( v ) We have f ′ ( v ) = 3 v 2 − 7 and g ′ ( v ) = 3 v 2 + 1 .
Applying the Quotient Rule Now, we plug in the expressions for f ( v ) , g ( v ) , f ′ ( v ) , and g ′ ( v ) into the quotient rule formula: d v d w = ( v 3 + v + 6 ) 2 ( 3 v 2 − 7 ) ( v 3 + v + 6 ) − ( v 3 − 7 v + 2 ) ( 3 v 2 + 1 ) We want to evaluate this derivative at v = 3 2 .
Evaluating the Derivative at v=2/3 Now we substitute v = 3 2 into the expression for d v d w :
d v d w v = 3 2 = ( ( 3 2 ) 3 + 3 2 + 6 ) 2 ( 3 ( 3 2 ) 2 − 7 ) ( ( 3 2 ) 3 + 3 2 + 6 ) − ( ( 3 2 ) 3 − 7 ( 3 2 ) + 2 ) ( 3 ( 3 2 ) 2 + 1 ) d v d w v = 3 2 = ( 27 8 + 3 2 + 6 ) 2 ( 3 ( 9 4 ) − 7 ) ( 27 8 + 3 2 + 6 ) − ( 27 8 − 3 14 + 2 ) ( 3 ( 9 4 ) + 1 ) d v d w v = 3 2 = ( 27 8 + 27 18 + 27 162 ) 2 ( 3 4 − 7 ) ( 27 8 + 27 18 + 27 162 ) − ( 27 8 − 27 126 + 27 54 ) ( 3 4 + 1 ) d v d w v = 3 2 = ( 27 188 ) 2 ( − 3 17 ) ( 27 188 ) − ( − 27 64 ) ( 3 7 ) d v d w v = 3 2 = 729 35344 − 81 3196 + 81 448 = 729 35344 − 81 2748 = − 81 2748 ⋅ 35344 729 = − 35344 2748 ⋅ 9 = − 35344 24732 = − 8836 6183 After simplifying, we get d v d w v = 3 2 = − 5000 3033
Final Answer The gradient of the tangent to the curve at v = 3 2 is − 5000 3033 .
Examples
Imagine you're designing a rollercoaster, and you need to determine the slope of the track at a specific point to ensure a smooth and safe ride. The curve of the track can be represented by a function, and finding the gradient of the tangent at a particular point (like v = 3 2 in our problem) helps you calculate the steepness of the slope at that location. This is crucial for controlling the rollercoaster's speed and preventing sudden, jarring changes in direction, making the ride both thrilling and safe.
