The Mean Value Theorem A student at the University of Connecticut - PowerPoint PPT Presentation

The Right Hand Limit: f ′ ( c ) = lim x → c + f ( x ) − f ( c ) . x − c Since f has a maximum at c , if x > c , then f ( x ) ≤ f ( c ), so f ( x ) − f ( c ) ≤ 0. But, if x > c , it’s also true that x − c > 0 and it follows that f ( x ) − f ( c ) ≤ 0. We see lim x → c + f ( x ) − f ( c ) is the x − c x − c limit of numbers less than or equal to 0 and therefore can’t be positive. It follows that f ′ ( c ) ≤ 0. Since f ′ ( c ) ≥ 0 and f ′ ( c ) ≤ 0, it follows that f ′ ( c ) = 0 QED Alan H. SteinUniversity of Connecticut

Consequences of Rolle’s Theorem Besides being a special case of the Mean Value Theorem and being a step in the path to proving the Mean Value Theorem, Rolle’s Theorem has some interesting applications of its own. Corollary A polynomial equation of degree n has at most n solutions. Alan H. SteinUniversity of Connecticut

Consequences of Rolle’s Theorem Besides being a special case of the Mean Value Theorem and being a step in the path to proving the Mean Value Theorem, Rolle’s Theorem has some interesting applications of its own. Corollary

Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial. Alan H. SteinUniversity of Connecticut

Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial. The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. Alan H. SteinUniversity of Connecticut

Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial. The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. The derivative of a quadratic polynomial is linear, having no more than 1 zero, so a quadratic can’t have more than 2 zeros. Alan H. SteinUniversity of Connecticut

Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial. The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. The derivative of a quadratic polynomial is linear, having no more than 1 zero, so a quadratic can’t have more than 2 zeros. The derivative of a cubic polynomial is quadratic, having no more than 2 zeros, so the cubic can’t have more than 3 zeros. Alan H. SteinUniversity of Connecticut

Since a polynomial may be differentiated as many times as necessary, with each derivative being a polynomial of lower degree, one immediate consequence of Rolle’s Theorem is that the derivative of a polynomial has at least one zero between each pair of distinct zeros of the original polynomial. The derivative of a linear polynomial is a non-zero constant, having no zeros, so a linear polynomial can’t have more than 1 zero. The derivative of a quadratic polynomial is linear, having no more than 1 zero, so a quadratic can’t have more than 2 zeros. The derivative of a cubic polynomial is quadratic, having no more than 2 zeros, so the cubic can’t have more than 3 zeros. This clearly goes on forever. The argument can be made rigorous through the use of Mathematical Induction . Alan H. SteinUniversity of Connecticut

The Mean Value Theorem Alan H. SteinUniversity of Connecticut

The Mean Value Theorem Theorem (The Mean Value Theorem) Suppose a function f is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) . Then there is a number c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ) or, equivalently, f ′ ( c ) = f ( b ) − f ( a ) . b − a Alan H. SteinUniversity of Connecticut

The Mean Value Theorem Theorem (The Mean Value Theorem) Suppose a function f is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) . Then there is a number c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ) or, equivalently, f ′ ( c ) = f ( b ) − f ( a ) . b − a Geometrically, the Mean Value Theorem says that if there is a smooth curve connecting two points, there must be some point in between at which the tangent line is parallel to the line connecting those two points. Alan H. SteinUniversity of Connecticut

The Mean Value Theorem Theorem (The Mean Value Theorem) Suppose a function f is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) . Then there is a number c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ) or, equivalently, f ′ ( c ) = f ( b ) − f ( a ) . b − a Geometrically, the Mean Value Theorem says that if there is a smooth curve connecting two points, there must be some point in between at which the tangent line is parallel to the line connecting those two points. Analytically, the Mean Value Theorem says the rate of change of a differentiable function must, at some point, take on its average, or mean value. Alan H. SteinUniversity of Connecticut

The proof of the Mean Value Theorem depends on the fact that the particular point at which the tangent line is parallel to the line connecting the endpoints also happens to be the point at which the curve is furthers away from that line. The proof essentially consists of applying Rolle’s Theorem to the function measuring the distance between the line and the curve. Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form y = f ( a ) + f ( b ) − f ( a ) · ( x − a ). b − a Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form y = f ( a ) + f ( b ) − f ( a ) · ( x − a ). b − a Since the second coordinate of a point on the curve with first coordinate x is f ( x ), the vertical distance between the line and the curve will equal Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form y = f ( a ) + f ( b ) − f ( a ) · ( x − a ). b − a Since the second coordinate of a point on the curve with first coordinate x is f ( x ), the vertical distance between the line and the curve will equal � � f ( a ) + f ( b ) − f ( a ) f ( x ) − · ( x − a ) = b − a Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form y = f ( a ) + f ( b ) − f ( a ) · ( x − a ). b − a Since the second coordinate of a point on the curve with first coordinate x is f ( x ), the vertical distance between the line and the curve will equal � � f ( a ) + f ( b ) − f ( a ) f ( x ) − · ( x − a ) = b − a f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a Alan H. SteinUniversity of Connecticut

Since the line goes between the points ( a , f ( a )) and ( b , f ( b )), its slope will be f ( b ) − f ( a ) and its equation may be written, in b − a point-slope form, y − f ( a ) = f ( b ) − f ( a ) · ( x − a ). b − a Solving for y , we may write the equation in the form y = f ( a ) + f ( b ) − f ( a ) · ( x − a ). b − a Since the second coordinate of a point on the curve with first coordinate x is f ( x ), the vertical distance between the line and the curve will equal � � f ( a ) + f ( b ) − f ( a ) f ( x ) − · ( x − a ) = b − a f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a We are now prepared to prove the Mean Value Theorem . Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a , b ]. Certainly, the fact that φ is both continuous and differentiable on [ a , b ] follows immediately from the fact that f is. In addition, Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a , b ]. Certainly, the fact that φ is both continuous and differentiable on [ a , b ] follows immediately from the fact that f is. In addition, φ ( a ) = f ( a ) − f ( a ) − f ( b ) − f ( a ) · ( a − a ) = 0 b − a Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a , b ]. Certainly, the fact that φ is both continuous and differentiable on [ a , b ] follows immediately from the fact that f is. In addition, φ ( a ) = f ( a ) − f ( a ) − f ( b ) − f ( a ) · ( a − a ) = 0 b − a and Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a , b ]. Certainly, the fact that φ is both continuous and differentiable on [ a , b ] follows immediately from the fact that f is. In addition, φ ( a ) = f ( a ) − f ( a ) − f ( b ) − f ( a ) · ( a − a ) = 0 b − a and φ ( b ) = f ( b ) − f ( a ) − f ( b ) − f ( a ) · ( b − a ) = b − a f ( b ) − f ( a ) − [ f ( b ) − f ( a )] = 0. Alan H. SteinUniversity of Connecticut

Proof: Let φ ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) · ( x − a ). b − a It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on the interval [ a , b ]. Certainly, the fact that φ is both continuous and differentiable on [ a , b ] follows immediately from the fact that f is. In addition, φ ( a ) = f ( a ) − f ( a ) − f ( b ) − f ( a ) · ( a − a ) = 0 b − a and φ ( b ) = f ( b ) − f ( a ) − f ( b ) − f ( a ) · ( b − a ) = b − a f ( b ) − f ( a ) − [ f ( b ) − f ( a )] = 0. Thus, there must be some c ∈ ( a , b ) such that φ ′ ( c ) = 0. We first obtain φ ′ ( c ) as follows. Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) b − a Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) b − a Since φ ′ ( c ) = 0, it follows that Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) b − a Since φ ′ ( c ) = 0, it follows that f ′ ( c ) − f ( b ) − f ( a ) = 0. b − a Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) b − a Since φ ′ ( c ) = 0, it follows that f ′ ( c ) − f ( b ) − f ( a ) = 0. b − a f ′ ( c ) = f ( b ) − f ( a ) b − a Alan H. SteinUniversity of Connecticut

φ ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a φ ′ ( c ) = f ′ ( c ) − f ( b ) − f ( a ) b − a Since φ ′ ( c ) = 0, it follows that f ′ ( c ) − f ( b ) − f ( a ) = 0. b − a f ′ ( c ) = f ( b ) − f ( a ) b − a f ′ ( c )( b − a ) = f ( b ) − f ( a ) QED Alan H. SteinUniversity of Connecticut

Consequences of the Mean Value Theorem Perhaps the most important consequence of the Mean Value Theorem is that it gives precise meaning to the most important single concept in elementary Calculus, Alan H. SteinUniversity of Connecticut

The Derivative Measures Rate of Change. Theorem a. If the derivative of a function is positive at all points on an interval, then the function is increasing on that interval. b. If the derivative of a function is negative at all points on an interval, then the function is decreasing on that interval. Alan H. SteinUniversity of Connecticut

To prove this, we need a precise definition of what it means to be increasing or decreasing. Alan H. SteinUniversity of Connecticut

To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition (Strictly Increasing) A function f is said to be strictly increasing on an open interval I if f ( a ) < f ( b ) whenever a , b ∈ I and a < b . Alan H. SteinUniversity of Connecticut

To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition (Strictly Increasing) A function f is said to be strictly increasing on an open interval I if f ( a ) < f ( b ) whenever a , b ∈ I and a < b . Definition (Nondecreasing) A function f is said to be nondecreasing on an open interval I if f ( a ) ≤ f ( b ) whenever a , b ∈ I and a < b . Alan H. SteinUniversity of Connecticut

To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition (Strictly Increasing) A function f is said to be strictly increasing on an open interval I if f ( a ) < f ( b ) whenever a , b ∈ I and a < b . Definition (Nondecreasing) A function f is said to be nondecreasing on an open interval I if f ( a ) ≤ f ( b ) whenever a , b ∈ I and a < b . Note the subtle difference. Often, we will simply say a function is increasing. Generally, in those cases, it will not really be important whether we mean strictly increasing or simply nondecreasing . Alan H. SteinUniversity of Connecticut

To prove this, we need a precise definition of what it means to be increasing or decreasing. Definition (Strictly Increasing) A function f is said to be strictly increasing on an open interval I if f ( a ) < f ( b ) whenever a , b ∈ I and a < b . Definition (Nondecreasing) A function f is said to be nondecreasing on an open interval I if f ( a ) ≤ f ( b ) whenever a , b ∈ I and a < b . Note the subtle difference. Often, we will simply say a function is increasing. Generally, in those cases, it will not really be important whether we mean strictly increasing or simply nondecreasing . There are similar definitions of the terms strictly decreasing and nonincreasing . Here, too, we will often use the ambiguous term decreasing . Alan H. SteinUniversity of Connecticut

Definition (Strictly Decreasing) A function f is said to be strictly decreasing on an open interval I if f ( a ) > f ( b ) whenever a , b ∈ I and a < b . Alan H. SteinUniversity of Connecticut

Definition (Strictly Decreasing) A function f is said to be strictly decreasing on an open interval I if f ( a ) > f ( b ) whenever a , b ∈ I and a < b . Definition (Nonincreasing) A function f is said to be nonincreasing on an open interval I if f ( a ) ≥ f ( b ) whenever a , b ∈ I and a < b . Alan H. SteinUniversity of Connecticut

Definition (Strictly Decreasing) A function f is said to be strictly decreasing on an open interval I if f ( a ) > f ( b ) whenever a , b ∈ I and a < b . Definition (Nonincreasing) A function f is said to be nonincreasing on an open interval I if f ( a ) ≥ f ( b ) whenever a , b ∈ I and a < b . With these definitions, we are ready to prove the derivative measures rate of change . We will prove just one of what are really four different parts, that if the derivative is strictly positive then the function is strictly increasing. Alan H. SteinUniversity of Connecticut

Proof. Suppose f ′ ( x ) > 0 ∀ x ∈ I and let a , b ∈ I , a < b . We need to prove that f ( a ) < f ( b ). Alan H. SteinUniversity of Connecticut

Proof. Suppose f ′ ( x ) > 0 ∀ x ∈ I and let a , b ∈ I , a < b . We need to prove that f ( a ) < f ( b ). Note: We’ve introduced the notation ∀ to mean for all or for every . Alan H. SteinUniversity of Connecticut

Proof. Suppose f ′ ( x ) > 0 ∀ x ∈ I and let a , b ∈ I , a < b . We need to prove that f ( a ) < f ( b ). Note: We’ve introduced the notation ∀ to mean for all or for every . Since f ′ ( x ) > 0 ∀ x ∈ I , it follows that f is continuous on [ a , b ] and differentiable on ( a , b ), so by the Mean Value Theorem there is some c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ). Alan H. SteinUniversity of Connecticut

Proof. Suppose f ′ ( x ) > 0 ∀ x ∈ I and let a , b ∈ I , a < b . We need to prove that f ( a ) < f ( b ). Note: We’ve introduced the notation ∀ to mean for all or for every . Since f ′ ( x ) > 0 ∀ x ∈ I , it follows that f is continuous on [ a , b ] and differentiable on ( a , b ), so by the Mean Value Theorem there is some c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ). Since f ′ ( x ) > 0 ∀ x ∈ I , it follows that f ′ ( c ) is positive. Since a < b , it follows that b − a is also positive, so that f ′ ( c )( b − a ) is also positive and hence f ( b ) − f ( a ) must be positive. Alan H. SteinUniversity of Connecticut

Proof. Suppose f ′ ( x ) > 0 ∀ x ∈ I and let a , b ∈ I , a < b . We need to prove that f ( a ) < f ( b ). Note: We’ve introduced the notation ∀ to mean for all or for every . Since f ′ ( x ) > 0 ∀ x ∈ I , it follows that f is continuous on [ a , b ] and differentiable on ( a , b ), so by the Mean Value Theorem there is some c ∈ ( a , b ) such that f ( b ) − f ( a ) = f ′ ( c )( b − a ). Since f ′ ( x ) > 0 ∀ x ∈ I , it follows that f ′ ( c ) is positive. Since a < b , it follows that b − a is also positive, so that f ′ ( c )( b − a ) is also positive and hence f ( b ) − f ( a ) must be positive. Clearly, f ( a ) must be smaller than f ( b ). Alan H. SteinUniversity of Connecticut

Very Different Functions Can’t Have the Same Derivative Alan H. SteinUniversity of Connecticut

Very Different Functions Can’t Have the Same Derivative It’s obvious that if two functions differ by only a constant term, then they will have the same derivative. Alan H. SteinUniversity of Connecticut

Very Different Functions Can’t Have the Same Derivative It’s obvious that if two functions differ by only a constant term, then they will have the same derivative. Example: d = d x 2 + 5 � x 2 � � � = 2 x . dx dx Alan H. SteinUniversity of Connecticut

Very Different Functions Can’t Have the Same Derivative It’s obvious that if two functions differ by only a constant term, then they will have the same derivative. Example: d = d x 2 + 5 � x 2 � � � = 2 x . dx dx A natural question is whether only functions differing by a constant can share the same derivative. The Mean Value Theorem enables us to see this is true. Alan H. SteinUniversity of Connecticut

Theorem If f ′ ( x ) = g ′ ( x ) for all x in some interval, then there is some constant k such that f ( x ) = g ( x ) + k. Alan H. SteinUniversity of Connecticut

Theorem If f ′ ( x ) = g ′ ( x ) for all x in some interval, then there is some constant k such that f ( x ) = g ( x ) + k. Proof: Let φ = f − g and let a be some fixed point in the interval. Now let x be in the interval. Clearly, the φ is both continuous and differentiable on the interval with endpoints a and x and we can apply the Mean Value Theorem . Alan H. SteinUniversity of Connecticut

Theorem If f ′ ( x ) = g ′ ( x ) for all x in some interval, then there is some constant k such that f ( x ) = g ( x ) + k. Proof: Let φ = f − g and let a be some fixed point in the interval. Now let x be in the interval. Clearly, the φ is both continuous and differentiable on the interval with endpoints a and x and we can apply the Mean Value Theorem . We use that language because it is possible that a < x and the interval is [ a , x ] but also possible that a > x and the interval is [ x , a ] . Alan H. SteinUniversity of Connecticut

By the Mean Value Theorem, there is some c in the interval such that φ ( x ) − φ ( a ) = φ ′ ( c )( x − a ). Alan H. SteinUniversity of Connecticut

By the Mean Value Theorem, there is some c in the interval such that φ ( x ) − φ ( a ) = φ ′ ( c )( x − a ). Because φ = f − g , it follows that φ ′ ( c ) = 0, so φ ( x ) − φ ( a ) = 0. Alan H. SteinUniversity of Connecticut

By the Mean Value Theorem, there is some c in the interval such that φ ( x ) − φ ( a ) = φ ′ ( c )( x − a ). Because φ = f − g , it follows that φ ′ ( c ) = 0, so φ ( x ) − φ ( a ) = 0. It follows that φ ( x ) = φ ( a ), or φ ( x ) = k , where we let k = φ ( a ). Alan H. SteinUniversity of Connecticut

By the Mean Value Theorem, there is some c in the interval such that φ ( x ) − φ ( a ) = φ ′ ( c )( x − a ). Because φ = f − g , it follows that φ ′ ( c ) = 0, so φ ( x ) − φ ( a ) = 0. It follows that φ ( x ) = φ ( a ), or φ ( x ) = k , where we let k = φ ( a ). Since φ = f − g , we have f ( x ) − g ( x ) = k , or f ( x ) = g ( x ) + k . QED Alan H. SteinUniversity of Connecticut

Corollary Only constant functions have derivatives which are identically 0. Alan H. SteinUniversity of Connecticut

Corollary Only constant functions have derivatives which are identically 0. This theorem will prove useful when we try to find functions with a given derivative. One application will be to determine the acceleration due to gravity. Alan H. SteinUniversity of Connecticut

Example An object is dropped from a height of 128 feet. How long does it take to reach the ground? Alan H. SteinUniversity of Connecticut

Example An object is dropped from a height of 128 feet. How long does it take to reach the ground? Solution Alan H. SteinUniversity of Connecticut

Example An object is dropped from a height of 128 feet. How long does it take to reach the ground? Solution Let: ◮ h be the height of the object, measured in feet. ◮ v be the velocity of the object, measured in feet per second. ◮ t be the time since the object was dropped, measured in seconds. Alan H. SteinUniversity of Connecticut

We Know: ◮ h = 128 when t = 0 ◮ v = 0 when t = 0 ◮ dh dt = v (Since velocity is the rate at which the height changes.) ◮ dv dt = − 32 (Since acceleration is the rate at which the velocity changes and the acceleration due to gravity is 32 feet per second per second and is in the downward, or negative, direction.) Alan H. SteinUniversity of Connecticut

frame ◮ The value of t when h = 0 Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that v = − 32 t + c for some constant c . Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that v = − 32 t + c for some constant c . Since v = 0 when t = 0, it follows that Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that v = − 32 t + c for some constant c . Since v = 0 when t = 0, it follows that 0 = − 32 · 0 + c Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that v = − 32 t + c for some constant c . Since v = 0 when t = 0, it follows that 0 = − 32 · 0 + c c = 0 Alan H. SteinUniversity of Connecticut

Since d dt ( − 32 t ) = − 32 and two functions can have the same derivative only if they differ by a constant, it follows that v = − 32 t + c for some constant c . Since v = 0 when t = 0, it follows that 0 = − 32 · 0 + c c = 0 v = − 32 t Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that h = − 16 t 2 + k for some constant k . Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that h = − 16 t 2 + k for some constant k . Since h = 128 when t = 0, it follows that Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that h = − 16 t 2 + k for some constant k . Since h = 128 when t = 0, it follows that 128 = − 16 · 0 2 + k Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that h = − 16 t 2 + k for some constant k . Since h = 128 when t = 0, it follows that 128 = − 16 · 0 2 + k k = 128 Alan H. SteinUniversity of Connecticut

Since d = − 32 t , dh � − 16 t 2 � dt = v = − 32 t and two functions can dt have the same derivative only if they differ by a constant, it follows that h = − 16 t 2 + k for some constant k . Since h = 128 when t = 0, it follows that 128 = − 16 · 0 2 + k k = 128 h = − 16 t 2 + 128 Alan H. SteinUniversity of Connecticut

Thus, when h = 0, Alan H. SteinUniversity of Connecticut