Calculus I Exam 2 Review Solutions 2019

Exam 2 Review Solutions 1. Finish the definition: f (x + h) − f (x) h (b) A function f is said to be differentiable at a point x = a if: f 0 (a) exists (as a limit) (a) The derivative of f is: f 0 (x) = lim h→0 (c) A function f is said to be differentiable on the (open) interval (a, b) if: f is differentiable at every point in the interval (a, b). 2. Short Answer: (a) How do we define the inverse sine function? (Pay attention to the domain, range and whether the domain, range are angle measures or the ratios of a triangle). SOLUTION: We first consider the domain and range of y = sin(x). Since this function is not 1-1 on its entire domain, we restrict the domain so that the restricted function is 1-1 (meaning the function passes the ”horizontal line test”). The standard restriction for the domain is below, together with the range: −π/2 ≤ x ≤ π/2 −1≤y ≤1 Now, the inverse sine is defined to be the inverse to the sine function, and its domain and range are: −1 ≤ x ≤ 1 − π/2 ≤ y ≤ π/2 √ −1 Extra: √ If we ask you to compute sin (1/ 2), this is the same as: Find angle θ so that sin(θ) = 1/ 2- However, θ must be only between −π/2 and π/2 (in this√case, π/4). You might compare this to the answer of something like: Find θ so that sin(θ) = 1/ 2, and θ ∈ [0, 2π]. In this case, you would also add the angle 3π/4. (b) What is a normal line? SOLUTION: The normal line to a function at a given point is the line that is perpendicular to the tangent line (at the same point). Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line. (c) How do we differentiate a function that involves the absolute value? SOLUTION: Rewrite the function into a piecewise form. For example,  x2 − 4 if x < −2 or x > 2 |x2 − 4| = −(x2 − 4) if − 2 ≤ x ≤ 2 sin(θ) =1 θ limθ→0 cos(θ)−1 =0 θ tan(3t) limθ→0 sin(2t) =? (Do (d) limθ→0 (e) (f) provide details) SOLUTION: We want to manipulate the expression so that we can use the two limits in (d) and (e): sin(3t) 1 sin(3t) 1 1 lim = lim · · = θ→0 cos(3t) sin(2t) θ→0 1 cos(3t) sin(2t) lim θ→0 3t sin(3t) 1 2t 3 3 · · · = ·1·1·1= 2t 3t cos(3t) sin(2t) 2 2 1 √ (g) If f (x) = x, find a formula for f 0 (x) using the definition of the derivative. SOLUTION: √ √ √ √ √ √ x+h− x x+h− x x+h+ x 0 f (x) = lim = lim ·√ √ = h→0 h→0 h h x+h+ x lim h→0 x+h−x 1 √ √ = √ 2 x h( x + h + x) NOTE: To get any credit, you must use the definition, not the power rule. (h) If f (x) = 3/x, find a formula for f 0 (x) using the definition of the derivative. SOLUTION: 3 −3 1 3x − 3x − 3h 3 f 0 (x) = lim x+h x = lim =− 2 h→0 h→0 h h x(x + h) x 3. Prove the Reciprocal Rule using the Product Rule (Hint: Start with f (x) = 1/g(x), then write f (x)g(x) = 1). SOLUTION: f (x)g(x) = 1 ⇒] f 0 (x)g(x)+f (x)g 0 (x) = 0 f 0 (x)g(x) = −f (x)g 0 (x) ⇒ ⇒ f 0 (x) = − f (x) g(x) g 0 (x) Now, substitute f (x) = 1/g(x) to get the final result: f 0 (x) = − g 0 (x) (g(x))2 4. Prove the Quotient Rule using the Product and Reciprocal Rules:   0 0  0  f (x) −g (x) 1 1 + f (x) = f (x) = f 0 (x) g(x) g(x) g(x) g(x))2 Now get a common denominator ((g(x))2 ) and simplify a bit: f 0 (x)g(x) − f (x)g 0 (x) (g(x))2 5. True or False, and explain: (a) The derivative of a polynomial is a polynomial. True. A polynomial is a function of the form a0 + a1 x + a2 x2 + . . . an xn , and its derivative will also have integer powers of x by the Power Rule, nxn−1 . p f 0 (x) d f (x) = √ (b) If f is differentiable, then dx 2 True: f (x) d d p 1 f 0 (x) f (x) = (f (x))1/2 = (f (x))−1/2 f 0 (x) = p dx dx 2 2 f (x) (c) The derivative of y = sec−1 (x) is the derivative of y = cos(x). False. The notation, sec−1 (x) is for the inverse secant function, which is not the reciprocal of the secant. For extra practice, to get the formula for the derivative of y = sec−1 (x): sec(y) = x 2 From this, draw a right triangle with one acute angle labelled y, the hypotenuse x and the adjacent √ length 1. This gives the length of the side opposite: x2 − 1. Now differentiate: sec(y) tan(y) From the triangle, sec(y) = x and tan(y) = √ dy =1 dx x2 − 1, so: dy 1 = √ dx x x2 − 1 (d) d x dx (10 ) = x10x−1 False. The Power Rule can only be used for xn , not ax . The derivative is 10x ln(10). (e) If y = ln |x|, then y 0 = x1 . TRUE. To see this, re-write the function:  ln(x) if x > 0 y= ln(−x) if x < 0 ⇒ y0 =  1 −x 1 x · (−1) if x > 0 if x < 0 From which we see that y 0 = x1 , x 6= 0. (f) The equation of the tangent line to y = x2 at (1, 1) is: y − 1 = 2x(x − 1) False. This is the equation of a parabola, not a line. The derivative, y 0 = 2x gives a formula for the slope of the tangent line, and is not the slope itself. To get the slope, we evaluate the derivative at x = 1, which gives y 0 = 2. The slope of the tangent line is therefore y − 1 = 2(x − 1). (g) If y = e2 , then y 0 = 2e False. e2 is a constant, so the derivative is zero. (h) If y = |x2 − x|, then y 0 = |2x − 1|. False. First, rewrite y, then differentiate:  x2 − x if x ≤ 0 or x ≥ 1 y= −(x2 − x) if 0 < x < 1 ⇒ y0 =  2x − 1 −(2x − 1) if x < 0 or x > 1 if 0 < x < 1 Compare this to |2x − 1|:  |2x − 1| = 2x − 1 −(2x − 1) if x ≥ 1/2 if x < 1/2 By the way, we also note that y is NOT differentiable at x = 0 or at x = 1 by checking to see what the derivatives are approaching as x → 0 and as x → 1. (i) If y = ax + b, then True. dy da dy da =x means that we treat a as an independent variable, and x, b as constants. 6. Find the equation of the tangent line to x3 + y 3 = 3xy at the point ( 23 , 23 ). We need to find the slope, dy dx x=3/2,y=3/2 3x2 + 3y 2 y 0 = 3y + 3xy 0 ⇒ y 0 (3y 2 − 3x) = 3y − 3x2 ⇒ y 0 = y − x2 y2 − x Substituting x = 3/2, y = 3/2 gives y 0 = −1, so the equation of the tangent line is y−3/2 = −1(x−3/2) 3 7. If f (0) = 0, and f 0 (0) = 2, find the derivative of f (f (f (f (x)))) at x = 0. A cute chain rule problem! Here we go- the derivative is: f 0 (f (f (f (x)))) · f 0 (f (f (x))) · f 0 (f (x)) · f 0 (x) Now subsitute x = 0 and evaluate: f 0 (f (f (f (0)))) · f 0 (f (f (0))) · f 0 (f (0)) · f 0 (0) = f 0 (0) · f 0 (0) · f 0 (0) · f 0 (0) = 24 = 16 8. If f (x) = 2x + ex , find the equation of the tangent line to the INVERSE of f at (1, 0). First, we verify that (0, 1) is on the graph of f : f (0) = 2 · 0 + e0 = 1 We know that, if f 0 (0) = m, then df −1 dx = x=1 1 m. Now, f 0 (x) = 2 + ex , so f 0 (0) = 3. Therefore, the slope of the tangent line to the inverse of f at x = 1 is 13 , and the equation is then: 1 1 1 (x − 1) or y = x − 3 3 3 y−0= 9. Derive the formula for the derivative of y = csc−1 (x) using implicit differentiation: SOLUTION: csc(y) = x, so that we can differentiate both sides: − csc(y) cot(y)y 0 = 1 ⇒ y0 = −1 csc(y) cot(y) Now, to simplify (and get a formula in terms of x), draw a triangle satisfying csc(y) = x- That is, label an angle y, then the length of the √ hypotenuse is x and the length of the side adjacent is 1. The length of the side opposite is therefore x2 − 1 (by the Pythagorean Theorem), and y0 = 10. Find the equation of the tangent line to −1 −1 = √ csc(y) cot(y) x x2 − 1 √ y + xy 2 = 5 at the point (4, 1). Implicit differentiation gives: 1 −1/2 0 y y + y 2 + 2xyy 0 = 0 2 Now we could solve for y 0 now, or substitute x = 4, y = 1: 1 −1/2 0 1 y + 12 + 2(4)(1)y 0 = 0 ⇒ y 0 = −2/17 2 The equation of the tangent line is y − 1 = 11. If s2 t + t3 = 1, find The notation dt ds dt ds and −2 17 (x − 4) ds dt . means that we are treating t as a function of s. Therefore, we have: 2st + s2 dt dt dt −2st + 3t2 =0⇒ = 2 ds ds ds s + 3t2 4 For the second part, we have two choices. One choice is to treat s as a function of t and differentiate: 2s ds ds −(s2 + 3t2 ) t + s2 + 3t2 = 0 ⇒ = dt dt 2st Another method is to realize that: ds 1 = dt = dt ds 1 =− −2st s2 +3t2 s2 + 3t2 2st Cool! 12. If y = x3 − 2 and x = 3z 2 + 5, then find We see that dy dz = dy dx dx dz , so we calculate dy dz . dy dx dx and dz : dy = 3x2 , dx so that dx = 6z dz dy = 3x2 · 6z = 3(3z 2 + 5)2 · 6z = 18z(3z 2 + 5)2 dz 13. A space traveler is moving from left to right along the curve y = x2 . When she shuts off the engines, she will go off along the tangent line at that point. At what point should she shut off the engines in order to reach the point (4, 15)? The unknown in the problem is a point on the parabola y = x2 . Let’s label that point as (a, a2 ). Now our goal is to find a. First, the line will go through both (a, a2 ) and (4, 15), so the slope will satisfy: m= a2 − 15 a−4 Secondly, the line will be a tangent line, so the slope will also be m = 2a. Equating these, we can solve for a: a2 − 15 = 2a ⇒ a2 − 15 = 2a(a − 4) ⇒ a2 − 8a + 15 = 0 ⇒ a = 3, a = 5 a−4 Since we’re moving from left to right, we would choose the smaller of these, a = 3. 14. A particle moves in the plane according to the law x = t2 + 2t, y = 2t3 − 6t. Find the slope of the tangent line when t = 0. The slope is dy dx , but we can only compute dx dt and dy = dx So, at t = 0, dy dx dy dt dx dt dy dt . = Note however, that 6t2 − 6 2t + 2 = −3. 15. Find h0 in terms of f 0 and g 0 , if: h(x) = SOLUTION: h0 (x) = f (x)g(x) f (x) + g(x) (f 0 g + f g 0 )(f + g) − (f g)(f 0 + g 0 ) f 2 g0 + f 0 g2 = (f + g)2 (f + g)2 16. The volume of a right circular cone is V = 13 πr2 h, where r is the radius of the base and h is the height. 5 (a) Find the rate of change of the volume with respect to the radius if the height is constant. SOLUTION: The question is asking for dV /dr, when h is constant: 2 dV = πrh dr 3 (b) Find the rate of change of the volume with respect to time if both the height and the radius are functions of time. SOLUTION: Now we want dV /dt if h, r are functions of time. That means the formula for V looks like: 1 V (t) = π(r(t))2 h(t) 3 So we’ll use the product rule (and factor the constant out):
π dV = 2r(t)r0 (t)h(t) + (r(t))2 h0 (t) dt 3 17. Find the coordinates of the point on the curve y = (x − 2)2 at which the tangent line is perpendicular to the line 2x − y + 2 = 0. First, recall that two slopes are perpendicular if they are negative reciprocals (like −3, 31 ). The slope of the given line is 2, so we want a slope of − 12 . The x that will provide this slope is found by differentiating: y 0 = 2(x − 2) ⇒ 2(x − 2) = − 7 1 1 ⇒ x = from which y = 2 4 16 18. For what value(s) of A, B, C does the polynomial y = Ax2 + Bx + C satisfy the differential equation: y 00 + y 0 − 2y = x2 As we did in class, compute the derivatives of y and substitute into the equation: y 0 = 2Ax + B, y 00 = 2A so that: 2A + 2Ax + B − 2(Ax2 + Bx + C) = x2 Now collect coefficients to get: (−2A)x2 + (2A − 2B)x + (B + 2A − 2C) = x2 for all x so the coefficient of x2 on both sides of the equation must be the same, −2A = 1 The coefficient of x on both sides of the equation must be the same, 2A − 2B = 0 And the constant terms must be the same, B + 2A − 2C = 0 This gives the solution, A = − 21 , B = − 21 , C = − 34 . 6 √ 19. If V = sin(w), w = u, u = t2 + 3t, compute: The rate of change of V with respect to w, the rate of change of V with respect to u, and the rate of change of V with respect to t. dV = cos(w) dw √ dV dV dw 1 cos( u) √ = · = cos(w) · √ = • du dw du 2 u 2 u • • √ dV dV dw du 1 cos( t2 + 3t) √ = · · = cos(w) · √ · (2t + 3) = · (2t + 3) dt dw du dt 2 u 2 t2 + 3t 20. Find all value(s) of k so that y = ekt satisfies the differential equation: y 00 − y 0 − 2y = 0. First, differentiate y, then substitute: y = ekt ⇒ y 0 = kekt y 00 = k 2 ekt ⇒ so that: k 2 ekt − kekt − 2ekt = 0 ⇒ ekt (k 2 − k − 2) = 0 Since ekt = 0 has no solution, the only solution(s) come from: k2 − k − 2 = 0 ⇒ (k + 1)(k − 2) = 0 so k = −1, k = 2 are the two values of k. 21. Find the points on the ellipse x2 + 2y 2 = 1 where the tangent line has slope 1. Implicit differentiation gives: 2x + 4yy 0 = 0 ⇒ y 0 = − To have a slope of 1 will mean that: −x 2y x 2y = 1, so that x = −2y. This means that any point on the ellipse satisfying x = −2y will have a tangent line with slope 1, so we look for the points of intersection between x = −2y and the ellipse x2 + 2y 2 = 1. Therefore, we substitute for x and get: (−2y)2 + 2y 2 = 1 ⇒ 6y 2 = 1 ⇒ 1 y = ±√ 6 To get the x− coordinate, you can either backsubstitute into the equation for the ellipse: r  2 ±1 2 2 x +2 √ =1 ⇒ x=± 3 6 √ Or you can use the fact that x = −2y, so in that case x = ∓2/ 6, or (− √26 , √16 ) or ( √26 , − √16 ) NOTE: These look like different values of x, but either is fine since: √ 2 2 2 √ =√ √ =√ 6 2 3 3 22. Differentiate. You may assume that y is a function of x, if not already defined explicitly. √ (a) y = log3 ( x + 1) Use the Chain Rule: 1 1 y0 = √ · √ ( x + 1) ln(3) 2 x 7 (b) √ 2xy + xy 3 = 5 (and solve for Before solving for y 0 , we get: dy dx ) 1 −1/2 (2xy) (2y + 2xy 0 ) + y 3 + 3xy 2 y 0 = 0 2 y0 = (c) y = −(y(2xy)−1/2 + y 3 ) x(2xy)−1/2 + 3xy 2 p x2 + sin(x)
−1/2 dy 1 2 (2x + cos(x)) = x + sin(x) dx 2 (d) y = ecos(x) + sin(5x ) y 0 = ecos(x) (− sin(x)) + cos(5x ) · 5x ln(5) (e) y = cot(3x2 + 5) y 0 = − csc2 (3x2 + 5)(6x) = −6x csc2 (3x2 + 5) (f) y = xcos(x) Use logarithmic differentiation: ln(y) = cos(x) · ln(x), so that 1 1 0 y = − sin(x) ln(x) + cos(x) · y x Multiply both sides of the equation by y, and back substitute y = xcos(x) to get:   cos(x) 0 cos(x) y =x − sin(x) ln(x) + x (g) y = p √ sin( x) y0 = (h) √ x+ √ 3 1 1 (sin(x1/2 ))−1/2 cos(x1/2 ) x−1/2 2 2 y=1 1 −1/2 1 −2/3 0 x + y y =0 2 3 y0 = − 3y 2/3 2x1/2 (i) x tan(y) = y − 1 tan(y) + x sec2 (y)y 0 = y 0 tan(y) = y0 1 − x sec2 (y) 8 √ 2 (j) y = x ex (x2 + 1)10 (Hint: Logarithmic Diff) First, we rewrite so that: √ 2 ln(y) = ln( x ex (x2 + 1)10 ) Use the rules of logarithms to re-write this as the sum: ln(y) = 1 1 ln(x) + x2 ln(e) + 10 ln(x2 + 1) = ln(x) + x2 + 10 ln(x2 + 1) 2 2 So far, we’ve only done algebra. Now it’s time to differentiate: 1 0 1 1 1 y = · + 2x + 10 · 2 · 2x y 2 x x +1 Simplifying, multiplying through by y: y0 = y  20x 1 + 2x + 2 2x x +1  Finally, back substitute y: y0 = √ 2 x ex (x2 + 1)10 ·  20x 1 + 2x + 2 2x x +1 
(k) y = sin−1 tan−1 (x) This is a composition, so use the chain rule: y0 = q (l) y = ln | csc(3x) + cot(3x)| Recall that the derivative of ln |x| is y0 = 1 x, 1 · 2
2 x +1 1 − tan−1 (x) 1 so using the Chain Rule:   1 · − csc(3x) cot(3x) · 3 − csc2 (3x) · 3 csc(3x) + cot(3x) Which can be simplified: y0 = (m) y = −2 √ 4 3 First, t −1/x −3 csc(3x)(cot(3x) + csc(3x)) = −3 csc(3x) csc(3x) + cot(3x) note that y = −2t−3/4 so y 0 = 23 t−7/4 (n) y = x3 y 0 = 3−1/x + x3−1/x ln(3) · x−2 √ (o) y = x tan−1 ( x) Overall, use the product rule (then a chain rule):   √ √ √ 1 1 x 0 −1 −1 y = tan ( x) + x · √ 2 · √ == tan ( x) + 2(x2 + 1) ( x) + 1 2 x ex (p) y = e2 Before putting in the values, note that this derivative will be in the form: y 0 = e( ) · d () d x (2 ) · e = e( ) · 2( ) ln(2) · ex dx dx Putting in the appropriate expressions gives us: ex x y 0 = e2 2e ln(2)ex 9 (q) Let a be a positive constant. y = xa + ax y 0 = axa−1 + ax ln(a) (r) xy = y x We must use logs first, since the exponents have x and y: ln(xy ) = ln(y x ) ⇒ y ln(x) = x ln(y) ⇒ y 0 ln(x) + y · Now isolate and solve for y 0 :   ln(y) − x y y 0 ln(x) − = ln(y) − ⇒ y 0 = y x ln(x) − (s) Rewrite first: y = ln y0 = 1 2  q 3x+2 3x−2  = 1 2 3 3 − 3x + 2 3x − 2 y x x y 1 1 = ln(y) + x · y 0 x y = y(x ln(y) − y) x(y ln(x) − x) (ln(3x + 2) − ln(3x − 2)) Now y 0 can be computed:  = 1 2  3(3x − 2) − 3(3x + 2) (3x + 2)(3x − 2)  = −6 (3x + 2)(3x − 2) 23. The solution is given below: 24. For the last problem, estimate the derivatives so that h0 (2) = f 0 (f (2))f 0 (2) = f 0 (1)f 0 (2) ≈ (−1)(−1) = 1 10