ERTH2020 Tutorial 1 Maths Revision Page 1 ERTH2020 Tutorial Exercise 1: Basic Maths Revision - Answers The purpose of this document is to provide answers - not working. Note that in an exam situation you should always show your working. Algebra 1. Consider the following: x = p (a + b) / c Rearrange so as to express b in terms of the other variables. Answer: b = xc/p -a 2. Suppose the variables in the above equation have values as follows: x = 10.0 p = 5.0 c = 14.2 a = -9.5 Answer: 37.9 What is the value of b? 3. One of the most important equations in exploration seismology is the so-called NMO equation, which describes the travel times of seismic waves reflected off subsurface interfaces. 2 2 tx = t0 + x 2 / v 2 In brief, the terms tx and t0 are travel times of seismic waves, x is the distance between seismic source and detector, and v is an average velocity down to the reflecting interface. (i) Suppose we have been able to measure the times tx and t0, and we know the distance x. We want to find the velocity v. Rearrange the above expression to allow this. Answer: v = x 2 / ( t x 2 - t 0 2 ))
ERTH2020 Tutorial 1 Maths Revision Page 2 (ii) If tx and t0 are 1.6s and 1.0s respectively, and the distance x is 5km, what is the average velocity (v) down to the reflector of interest. Answer: v = 4003 m/s 4. In electrical geophysics the apparent resistivity for the Wenner electrode arrangement is given by = 2 a V / I In a particular survey the electrode spacing (a) is 20.0m. The measured voltage (V) and current (I) are 45 mv and 120mA respectively. What is the apparent resistivity of the rocks in this area? What are the units? Answer: = 47.1 m 5. An electrical resistivity survey is being carried out over a homogenous granite with a resistivity of approximately 1000 m. In this case the electrode spacing (a) is 10m. The generator being used supplies a current of 0.2 amps. What would be the voltage measured between the potential electrodes. Give the answer in volts and in millivolts. Answer: V = 3.183V = 3183mV 6. An important equation used in geophysics is the 2-layered seismic refraction equation: t = x/v 2 + 2Z cos i c / v 1 This is commonly used to analyse weathering profiles (e.g. to find depth to the base of weathering.) (i) On a plot of t vs x, what is the slope of the line, and what is the intercept on the t axis. Answer: slope = 1/ v 2 intercept = 2Z cos i c / v 1 (ii) On a particular seismic record, we measure the intercept as 0.02 s. The near-surface velocity (v 1 ) has been measured as 750 m/s. Assume that the trigonometric factor cos i c can be approximated as 1. Solve for the depth of the refracting interface (Z), which in this case is the base of weathering. Answer: 7.5m
ERTH2020 Tutorial 1 Maths Revision Page 3 Trigonometry Right-angled triangles appear in many geophysical developments. They have a number of interesting properties. The lengths of the sides are related via Pythagoras' Theorem. The basic trig functions (sin, cos, tan) represent ratios of sides of a right-angle triangle, and are used for convenience in a range of mathematical problems. 7. In each of the following right-angled triangles deduce the length of the third side, and compute sin, cos and tan for the marked angle, without using the trig functions on your calculator. 25.7 957.2 14.6 1234.5 Answers: left triangle l = 29.6 sin cos tan Answers: right triangle l = 779.6 sin cos tan
ERTH2020 Tutorial 1 Maths Revision Page 4 8. With reference to the right-angled triangle below, write down the expressions for sin cos and tan p q r Show that tan sin cos Using Pythagoras' Theorem show that sin 2 cos 2 Answers: (a) sin p/q cos r/q tan p/r (b) tan p/r = q sin q cos = sin cos (c) sin 2 cos 2 p 2 /q 2 r 2 /q 2 = (p 2 r 2 ) / q 2 = q 2 / q 2 (pythagoras) = 1 In the right-angled triangles below deduce the lengths a, b, c. 19.5 40 b 65 4.27 124.35 30 c a Answers: a = 7.40 b = 79.93 c = 46.14
ERTH2020 Tutorial 1 Maths Revision Page 5 10. In most practical problems we will measure angles in degrees. Another useful measure of angles is the radian. When using a calculator we need to be careful which mode we are in. We also need to be able to convert from one unit to the other. (i) Express the following angles in radians (a) 180 (b) 90 ( c ) 23.5 (ii) Express the following angles in degrees (a) 3.14159 radians (b) 6.26 radians (c ) -0.45 radians Answers: (i) 3.14159, 1.5708, 0.4102 (ii) 180, 358.67, - 25.78 11. The following diagram shows the path taken by a seismic wave reflecting off a horizontal rock interface below the surface. The geophone detector (G) is at a horizontal distance of 1500m from the sound source (S). We are able to deduce that the angles of incidence and reflection are each 17 as marked. Use trigonometry to find the depth of the reflector. S 1500m G 17 17 Answer: Depth of reflector = 2453 m
ERTH2020 Tutorial 1 Maths Revision Page 6 12. The following diagram shows a weathered layer (30m thick) overlying bedrock. Seismic waves travel at speeds of v W =1000 m/s and v B =3000 m/s respectively in these layers. The dashed line represents a direct wave travelling from source (S) to geophone (G) just below the surface. The solid line represents a refracted wave from S to G. This wave dives into the bedrock, skims just below the bedrock interface, and then returns to the surface. The critical angle (i C ) made with the normal obeys the relationship sin i C = v W / v B (i) Show that the travel time of the direct wave is 0.1s. (ii) Show that the critical angle i C is 19.47 degrees. (iii) Show that the travel time of the refracted wave is 0.09s (iv) Which wave (direct or refracted) arrives at the geophone first? S 100m G * v W =1000 m/s i C i C 30m v B =3000 m/s Answers: (i) t D = x / v W = 100./1000. = 0.1 s (ii) i C = sin -1 (1000./3000.) = 19.47 degrees (iii) Break up path into 3 segments, use trig to find segment lengths, add the times. (iv) refracted wave arrives first, even though it travels further.
ERTH2020 Tutorial 1 Maths Revision Page 7 Calculus: Differentiation 13. Consider the following plot of the simple function y = 2.5 x -15 (i) Suppose we want to find the slope of the curve at x = 5.0. We could do this 'from first principles' by calculating the function values at x=4.9 and at x=5.1, and then using slope = (y2-y1) / (x2-x1). What slope do you get? Answer: For x1=4.9, y1= -2.75. For x2=5.1, y2= -2.25 y/ x = 0.5/0.2 = 2.5 (ii) Check this result by finding the derivative dy/dx and evaluating it at x=5.0 Answer: dy/dx = 2.5 (same for all values of x).
ERTH2020 Tutorial 1 Maths Revision Page 8 14. Consider the following plot of a simple quadratic function y = x 2 + 20 (i) Estimate the slopes of the curve, from first principles, at x= -5 and x = 7.5 [method as in Q13(i)] Answer: Slope at x = -5 For x1=-5.1, y1= 46.01 For x2=-4.9, y2= 44.01 y/ x = -2.0/0.2 = -10.0 Slope at x = 7.5 For x1=7.4, y1= 74.76 For x2=7.6, y2= 47.76 y/ x = 3.0/0.2 = 15.0
ERTH2020 Tutorial 1 Maths Revision Page 9 (ii) Estimate the slopes at x= -5 and x = 7.5 using calculus. Answer: dy/dx = 2x At x = -5, dy/dx = -10.0 At x = 7.5, dy/dx = 15.0 (iii) Which approach is more accurate? Answer: Calculus (iv) Which approach is easier? Answer: Calculus
ERTH2020 Tutorial 1 Maths Revision Page 10 Calculus: Integration 15. Refer back to the linear function given in Q13. (i) By analysing triangles, estimate the total area under the curve in the region between x=-5 and x=10. Note that by convention, area above the y=0 axis is taken as positive and area below the axis is taken as negative. In this case some of the area will be negative, and some will be positive. Answer: Negative area = -151.25, Positive area = 20.0 TOTAL AREA = -131.25 (ii) Write down the indefinite integral of the function y = 2.5 x -15 Answer: 1.25 x 2-1 5 x +C (iii) Hence write down the definite integral for the function over the range x=-5 and x=10 Answer: [1.25 x 2-1 5 x ] -5 10 (iv) Evaluate this definite integral and compare your answer to (i) Answer: -131.25 (Agrees with first principles)
ERTH2020 Tutorial 1 Maths Revision Page 11 16. Consider the quadratic function plotted above (Q14). (i) By approximating the curve with straight lines, and analysing triangles, estimate the area under the curve for the region x=-10 to x=5. Answer: Using 3 triangles (x=-10,-5,0,5) Estimated Area = 412.5 + 162.5 + 162.5 = 737.5 (Because of the way the triangles are built this will overestimate the true area.) (ii) Compute the area under the curve using calculus. Answer: [x 3 / 3 + 20 x ] -10 5 = 675.0 (iii) Which approach is easier? (iv) Which approach is more accurate? Answer: calculus Answer: calculus
ERTH2020 Tutorial 1 Maths Revision Page 12 Calculus: Practical Applications 17. I have arrived at the gate to the beach (Point G on the accompanying plan-view diagram) and I want to walk to Point S, 100m along the beach, where the surf looks good. On the soft sand I can walk at 2.5 km/hr. On the hard sand (right at the waters edge) I can walk at 5 km/hr. I want to get to the surf as quickly as possible! 100 m Water S P x Hard sand (v H =5km/hr) 50m Soft sand (v S =2.5km/hr) Fence G (i) Use calculus to show that the take-off direction ( ) which will give the shortest possible time is 30 degrees, and that the time taken is then approximately 134.4 seconds. [Method 1: Express the total time as a function of x, minimise with respect to x, and finally find Outline of Answer: Note: Take care with units. t = (x 2 +2500) 0.5 / vs + (100 -x) / v H dt/dx = x (x 2 +2500) -0.5 / v S - 1 / v H Set dt/dx = 0, yields x = 28.87 m hence atan (28.87 / 50) = 30.0 degrees Substitute x=28.87 into time equation to obtain the minimum time t =134.4 s.
ERTH2020 Tutorial 1 Maths Revision Page 13 (ii) Repeat Part (i) but now derive your time equation directly as a function of and then differentiate with respect to Outline of Answer: t = 50 / v S cos (100-50 tan ) / v H dt/d 50 sec tan / v S - 50 sec 2 / v H Set dt/d 0 50 tan / v S - 50 sec / v H = 0 sin = v S / v H degrees (iii) Use a numerical approach to confirm your optimum value. [ Method: Using excel or a programming language such as Python plot the total time from G to P to S as a function of Read off the optimum value, and the corresponding travel time.] Outline of Answer: Plot this function: t = 50 / v S cos (100-50 tan ) / v H and pick off the angle which minimises the time. (iv) Which approach (analytical or numerical) do you prefer? Answer: Often, numerical will be quicker, but analytical is also useful to confirm that your code is correct.